ELECTRICAL 
CHARAGTERISTICS 

of 

Transmission  Circuits 


<r      0  .        I 


THE  LIBRARY 

OF 

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OF  CALIFORNIA 

DAVIS 

FROM  THE  LIBRARY 

OF 

SOPHIA  L.  MC  DONALD 


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ELECTRICAL 
CHARACTERISTICS  OF 
TRANSMISSION  CIRCUITS 


REPRINT    82 
FEBRUARY.  1922 


w 

(WCSTINCHOUSCA 
ELECTRIC 


Reprinted  from  artidea  originally  appearing  in  the  Electric  Journal 


WESTINGHOUSE  ELECTRIC  &  MANUFACTURING  CO. 

EAST  PITTSBURGH,  PA. 


LIBRARY 


Copyrighted,  1922, 

BY    THE 

Westinghouse  Electric  &  Manufacturing  Co. 
East  Pittsburgh,  Pa. 


PREFACE 


THE  rapid  expansion  in  distributing  and  transmission  systems  will 
continue  unabated  until  the  natural  power  resources  will  have  been 
fully  developed.  This  expansion  will  necessitate  a  tremendous 
amount  of  arithmetical  labor  m  connection  with  the  proper  solution  and 
calculation  of  performance  of  projected  transmission  and  distributing 
circuits.  It  will  demand  much  valuable  time  and  energy  in  the  educa- 
tion of  the  younger  engineers  now  going  thru  the  technical  schools  and 
others  who  will  follow  them.  It  was  primarily  to  assist  these  younger 
engineers  by  making  their  work  more  easy  and  less  liable  to  error,  and 
providing  them  with  all  necessary  tools  that  the  data  in  this  book  have 
been  compiled. 

Many  articles  each  pertaining  to  some  particular  method  of  solution 
of  transmission  circuits  have  been  published  from  time  to  time.  This  book 
constitutes  a  review  of  each  of  numerous  methods  perviously  proposed  by 
different  authors  with  examples  illustrating  each  method  of  solution  and 
the  accuracy  which  may  be  expected  by  its  use.  Thus  by  permission  of 
various  authors  the  reader  of  this  book  is  provided  with  a  choice  of  nu- 
merous methods  ranging  between  the  most  simplified  graphical  forms  of 
solutions  and  complete  mathematical  solutions.  He  is  also  provided  with 
ntunerous  and  extensive  tables  of  circviit  and  other  constants  making  it 
uimecessary  for  him  to  lose  time  and  risk  making  mistakes  in  calculating 
constants  for  each  case  in  question.  Much  effort  has  been  expended  with 
a  view  of  simplifying  explanations  by  the  aid  of  supplementary  diagrams 
and  tabulations.  The  engineer  upon  whose  lot  it  only  occasionally  falls 
to  determine  the  size  of  conductors  and  performances  of  circuits  appreciates 
how  easy  it  is  to  make  errors  in  calculations  which  may  prove  very  serious 
and  should  find  the  quick  estimating  tables  very  useful  particularly  for 
short  line  solutions. 

For  those  preferring  to  avoid  the  more  mathematical  solutions  the  all 
graphical  methods  for  solving  long  line  problems  including  the  Wilkinson 
&  Kennelly  charts  for  obtaining  graphically  the  auxiliary  constants  should 
prove  helpful. 

When  borrowed  material  has  been  used  in  this  book  full  credit  has 
been  given  the  author  at  the  place  the  material  is  used.  It  is  desired, 
however,  at  this  place  to  mention  the  high  appreciation  of  assistance  given 
by  Ralph  W.  Atkinson,  Herbert  B.  Dwight,  Dr.  A.  E.  Kennelly,  Dr.  A.  S. 
McAllister,  Ralph  D.  Mershon,  F.  W.  Peak  Jr.,  J.  F.  Peters,  Charles  R. 
Riker  and  T.  A.  Wilkinson. 


y^  nuox 


CONTENTS 


Page 

Chapter  I            Resistance,  Skin  Effect,  Inductance 1 

Chapter  II          Reactance,  Capacitance,  Charging  Current 10 

Chapter  III         Quick  Estimating  Tables 23 

Chapter  IV         Corona  Effect 35 

Chapter  V           Speed  of  Electric  Propagation,  Resonance,  Parallel- 
ing Transmission  Circuits,  Heating  of   Bare  Con- 
ductors out   of  Doors 40 

Chapter  VI         Determination  of  Frequency  and  Voltage 45 

Chapter  VII       Performance  of  Short  Transmission  Lines 49 

Chapter  VIII      Performance  of  Long  Transmission  Lines  (graphical)  61 

Chapter  IX         Performance  of  Long   Transmission   Lines  (mathe- 
matical)    77 

Chapter  X          Hyperbolic  Functions 88 

Chapter  XI         Performance  of  Long  Transmission  Lines  (by  hyper- 
bolic functions) 95 

Chapter  XII       Comparison  of  Various  Methods Ill 

Chapter  XIII     Cable  Characteristics 121 

Chapter  XIV      Synchronous    Motors    and    Condensers    for    Power 

Factor  Improvement 129 

Chapter  XV       Phase  Modifiers  for  Voltage  Control 138 

Chapter  XVI     A  Typical  220  Kv.  Problem 145 


ELECTRICAL  CHARACTERISTICS 
of  TRANSMISSION  CIRCUITS 


CHAPTER  I 
RESISTANCE-SKIN  EFFECT-INDUCTANCE 


THE  transmission  of  alternating-current  power  in- 
volves three  separate  circuits,  one  of  which  is 
composed  of  the  wires  forming  the  transmission 
line,  while  the  others  lie  in  the  medium  surrounding  the 
wires.  The  constants  of  these  circuits  are  interde- 
pendent; although  any  one  may  vary  greatly  from  the 
others  in  magnitude.*  There  is  first  the  electric  cir- 
cuit through  the  conductors.  Then  since  all  magnetic 
and  dielectric  lines  of  force  are  closed  upon  themselves 
forming  complete  circuits  there  is  a  magnetic  and  a 
dielectric  circuit.  The  magnetic  circuit  consists  of 
magnetic  lines  of  force  encircling  the  current  carrying 
conductors  and  the  dielectric  circuit  the  dielectric  lines 
of  force  terminating  in  the  current  carrying  conductors. 
The  close  analogy  of  these  is  given  in  Table  A,  a  care- 
ful study  of  which  will  help  those  not  familiar  with  the 
subject  to  a  clearer  understanding  of  what  happens  in 
an  alternating-current  transmission  circtiit. 

»  For  a  unidirectional  constant  current  the  magnetic 
field  remains  constant,  and  similarly  for  a  unidirectional 
constant  voltage  the  dielectric  field  is  constant.  With 
both  the  current  and  the  voltage  unidirectional  and  con- 
stant, the  electric  circuit  alone  enters  into  the  calcula- 
tions. A  changing  magnetic  flux  introduces  a  voltage 
into  the  electric  circuit  which  modifies  the  initial  or 
impressed  voltage.  This  eflfect  of  the  magnetic  circuit, 
which  is  measured  by  the  inductance  L,  storing  the 
energy  o.^i^L,  is  a  function  of  the  current,  and  hence 
is  of  most  importance  in  dealing  with  heavy  current 
circuits.     Similarly  a  changing  electrostatic   flux  adds 


*For  a  further  description  of  these  circuits  see  "Alteniat- 
iiig  Currents"  by  Prof.  Carl  E.  Magnusson,  from  which  Figs. 
I  to  5  are  reproduced  with  the  permission  of  the  author. 


(vectorially)  a  current  to  the  main  power  current. 
This  effect  of  the  dielectric  circuit,  which  is  measured 
by  the  capacitance,  storing  the  energy  o.^e^C,  is  a  func- 
tion of  the  voltage,  and  hence  is  of  most  importance  In 
dealing  with  high-voltage  circuits. 

In  an  alternating-current  circuit,  both  the  voltage 
and  the  current  are  continually  varying  in  magnitude, 
and  morever,  reversing  in  direction  for  each  successive 
half  cycle.  Therefore,  with  alternating  currents, 
energy  changes  occur  continuously  and  simultaneously 
in  the  interlinked  magnetic,  dielectric  and  electric  cir- 
cuits. 

Figs.  I  to  5  inclusive  illustrate  the  magnetic  and 
dielectric  field  surrounding  conductors  carrying  current. 
Figs.  I  and  3  represent  respectively  the  magnetic  and 
dielectric  circuits  when  the  conductors  are  far  apa'-t 
and  Figs.  2  and  4  when  they  are  close  together.  Fig. 
5  represents  the  resultant  of  the  superimposed  mag- 
netic and  dielectric  fields. 

The  magnetic  field  surrounding  a  conductor  which 
is  not  influenced  by  any  other  field  is  represented  by 
concentric  circles.  This  field  is  strongest  at  the  surface 
of  the  conductors  and  rapidly  decreases  with  increasing 
distance  from  the  conductor  as  indicated  by  the  spac- 
ing of  the  lines  of  Figs,  i  and  2. 

The  dielectric  stresses  surrounding  conductors  are 
represented  by  lines  drawn  radially  from  the  con- 
ductor. The  strength  of  the  dielectric  field  likewise 
decreases  with  the  distance  from  the  conductor  as  is 
indicated  by  the  widening  of  the  space  between  the 
lines.  The  mignetic  and  the  dielectric  lines  of  force 
always  cross  each  other  at  right  angles,  as  shown  in 
Fig-  5- 


RESISTANCE—SKIN  EFFECT— INDUCTANCE 


RESISTANCE  OF  COPPER  CONDUCTORS 
In  Table  I  the  resistance  per  thousand  feet  is  listed 
and  in  Table  II  per  mile  of  single  conductor.  Values 
are  given  for  both  solid  and  stranded  copper  conductors 
at  both  loo  and  97.3  percent  conductivity  and  corres- 
ponding to  various  temperatures  between  zero  and  75 
degrees  C.  The  foot  notes  with  these  tables  cover  all 
of  the  pertinent  data  upon  which  the  values  are  based. 
The  resistance  values  in  Table  I  corresponding  to 
temperatures  of  25  and  65  degrees  C.  were  taken  from 


FIG  I  f  IG  '' 

MAGNETIC  FIELD  OF  SINGLE  CONDUCTOR  [  MAGNETIC  FIELD  Of  CIRCUIT 

THE    MAGNETIC    CIRCUIT 


FIG.  3 
DIELECTRIC  FIELD  OF  SINGLE  CONDUCTOR  DIELECTRIC  FIELD  OF  CmCUIT 

THE    DIELECTRIC    CIRCUIT' 


FIG.  6,  COMBINED  DIELECTRIC  AND  MAGNETIC  CIRCUITP 


THE    ELECTRIC    CIRCUIT 

Bulletin  31  of  the  Bureau  of  Standards  issued  April  ist, 

1912.     The  resistance  values   (taking  into  account  the 

expansion  of  the  metal  with  rise  in  temperature)  for  the 

other  temperatures  were  calculated  in  accordance  with 

the  following  rule  from  page  10  of  Bulletin  No.  31. 

The  change  of  resistivity  of  copper  per  degree  C.  is 
a  constant,  independent  of  the  temperature  of  reference 
and  of  the  sample  of  copper.  This  resistivity-temperature 
constant  may  be  taken  for  general  purposes  as  0.0409  ohm 
(mil  foot). 

As  an  illustration : — A  2  000  000  circ.  mil  stranded 

copper  conductor  at    100  percent   conductivity,   has   .1 

resistance  of  0.00623  ohm  per  1000  feet  at  65  degrees  C. 

Required  to  calculate  its  resistance  at  zero  degrees  C. 


65X0.0409  =  2.6585  ohms  (mil-foot)  temper- 
ature   correction    or    2658.5    ohms    (mil,    1000    feet). 

2658.5 

=  0.0013^  ohm  change  m  resistance.  0.00623 

2000000 

— 0.00133  =  0.0049  ohm  resistance  at  zero  degrees  C. 
It  has  been  customary  to  publish  tables  of  resist- 
ance values  based  upon  a  temperature  of  20  degrees  C. 
and  100  percent  conductivity.  The  operating  tempera- 
tures of  conductors  carrying  current  is  usually  con- 
siderably higher  than  20  degrees  C.  and  therefore  cal- 
culations based  upon  this  temperature  do  not  often  re- 
present operating  conditions.  Neither  does  copper  of 
100  percent  conductivity  represent  the  usual  condition 
for  transmission  circuit  copper,  whose  average  conduct- 
ivity is  probably  nearer  97.3  percent.  The  values  in 
Tables  I  and  II  furnish  a  comparison  of  resistance  for 
annealed  and  hard  drawn  copper  of  stranded  and  solid 
conductors  at  various  temperatures  based  upon  the  new 
"Annealed  Copper  Standard". 

SKIN   EFFECT 

A  solid  conductor  may  be  considered  as  made  up 
of  separate  filaments,  just  as  a  piece  of  wood  is  made 
up  of  separate  fibres.  As  a  stranded  conductor  is 
actually  made  up  of  a  number  of  separate  wires,  such 
a  conductor  will  be  considered  in  the  following  ex- 
planation. The  inductance  of  the  various  wires  of  the 
cable  will  be  different,  due  to  the  fact  that  those  wires 
near  the  center  of  the  cable  will  be  linked  by  more  flux 
lines  than  are  the  wires  near  the  outer  surface.  The 
self-induced  back  e.m.f.  will  therefore  be  greater  in  the 
wires  located  near  the  center  of  the  cable.  The  higher 
reactance  of  the  inner  wires  causes  the  current  to  dis- 
tribute in  such  a  manner  that  the  current  density  will 
be  less  in  the  interior  than  at  the  surface.  This  crowd- 
ing of  the  current  to  the  surface  or  "skin"  of  the  wire 
is  known  as  "skin  effect". 

Since  the  self-induced  e.m.f.  is  proportional  to  the 
frequency  as  well  as  to  the  total  flux  linked,  the  skin 
effect  becomes  more  pronounced  at  higher  frequencies 
of  the  impressed  e.m.f.  It  also  becomes  greater  the 
larger  the  cross-section,  the  greater  the  conductivity 
and  the  greater  the  permeability  of  the  conductor. 

As  a  result  the  effective  resistance  of  a  conductor 
to  alternating  current  is  greater  than  to  direct  current. 
The  effective  resistance  of  nonmagnetic  conductors  to 
alternating  current  may  be  obtained  by  increasing  their 
direct-current  resistances  by  the  percentages  in  Table 
B,  which  were  derived  by  the  formulas  in  Pender's 
Handbook.  Thus  the  ohmic  resistance  of  a  1000  000 
circ.  mil  cable  is  approximately  8.4  percent  greater  at 
60  cycles  than  its  resistance  to  direct  current  at  a  tem- 
perature of  25°C.  If  the  temperature  of  the  conductor 
is  65  °C,  its  60  cycle  ohmic  resistance  will  be  approxi- 
mately 6.4  percent  greater  than  its  direct-current  re- 
sistance. The  practical  result  of  skin  effect  is  to  re- 
duce the  carrying  capacity  of  large  cables.  As  in- 
dicated by  the  values  in  Table  B,  skin  effect  may  be 
neglected  when  employing  non-magnetic  conductors  ex- 


RESISTANCE—SKIN  EFFECT— INDUCTANCE 


cept  in  the  use  ot  very  large  diameters.     It  is  usual  to  viently  large,  a  thousandth  part  of  it,  called  the  miUi- 

manufacture  cables  of  very  large  diameter,  especially  henry,  is  the  usual  practical  unit.     This  unit  is  the  co- 

for  service  at  high  frequencies,  with  a  non-conducting  efficient   of   self-induction   and   is   represented   by   the 

core.  In  case  of  magnetic  conductors,  such  as  steel  wire  letter  L. 


or  cable,  as  is  some  times  used  for  long  spans  or  short 
high  voltage  feeders,  skin  effect  must  be  carefully  con- 
sidered.* 


DISTRIBUTION  OF  FLUX 

When  current  flows  through  a  conductor,  a  mag- 
netomotive  force    (m.m.f.)    is  established   of  a   value 


TABLE  A— COMPARISON  OF  THE  THREE  CIRCUITS 


The  Electric  Circuit 


The  Magnetic  Circuit 


The  Dielectric  Circuit 


Current    / 
Voltage    E=RI 
Electric  Power 


Resistivity 
Resistance  R—W/P 


Magnetic  Flux    <t> 
Magnetomotive  Force    /•"=»! 
Magnetic  Energy 


Dielectric  Flux    V 
Electromotive  Force  E-=.QIC 
Dielectric  Energy 


Reluctivity 
Reluctance    R 
Inductance  Ij^^li 


Reactance  .r=  wL — \l  uiC 


Elastivity   \IK 
Elastance    5" 
Capacitance  C=V  IE 


-  Impedance  z  =  1/  r'  -J-  .^^ 


Conductivity  -y 


Conductance 


I  g=W/E 
\  9=rlc 


Permeability  ix  =  BIH 
Permeance  jW=  <^  /4JrF 


Admittance  y=i / :s:ig^^  J  b  =  \/  g'  -f-  b'-- 


INDUCTANCE 

Any  moving  mass,  for  instance  a  flywheel  in 
motion,  will  resist  a  change  in  velocity.  That  is,  the 
inertia  of  the  moving  mass  will  tend  to  keep  the  mass 
moving  when  disconnected  from  the  source  of  power. 
On  the  other  hand  the  inertia  will  oppose  any  effort 
to  speed  up  the  movement  of  the  mass. 

In  a  similar  manner,  the  inductance  of  an  electric 
circuit  resists  a  change  in  current.  The  cause  of  in- 
ductance in  an  electric  circuit  is  the  magnetic  field  which 
surrounds  the  circuit.  When  the  current  changes  this 
magnetic  field  changes  correspond- 
ingly,  and  in   effect   cuts  the  con-  TABLE  B— INCREASE  OF  EFFECTIVE  RESISTANCE  DUE  TO  SKIN  EFFECT. 


proportional  to  the  current. 
This    m.nxf.     is    of    zero 
value  at  the  center  of  the 
conductor  and  increases  as 
the  square  of  the  distance 
from  the   center  until   the 
surface  is  reached,  i  (This 
statement  as  well  as  those 
following  is  based  upon  the 
assumption   of   a    uniform 
distribution  of   current 
throughout   the   conductor, 
the     conductor     being     of 
non-magnetic  material  and 
located     i  n     non-magnetic 
surroundings,  such  as  air).     At  the  surface  it  becomes 
maximum  for  a  given  current  and  remains  at  this  maxi- 
mum value  for  all  distances  beyond  the  surface.     It  is 
customary  to  think  of  the  magnetic  field  surrounding 
conductors  as  concentric  circles  of  lines  of  force. 

A  physical  picture  of  the  magnetic  field  density 
surrounding  a  current  carrying  conductor  A  is  shown 
by  Chart  I.  The  magnetic  density  due  to  the  return 
circuit  (conductor  B)  is  indicated  in  outline  by  broken 
lines.  The  horizontal  divisions  represent  the  distance 
from  the  center  of  conductor  A  and  the  height  of  the 


I     Permittivity    A' 

Permittance 
I     (Capacitance)     C 
Susceptance  b=x/:r 


ductor,  producing  an  e.m.f.  in  it. 
This  e.m.f.  of  self  induction  has 
such  a  direction  as  to  resist  the 
change  in  current.  While  the  cur- 
rent is  increasing,  energy  is  stored 
in  the  magnetic  field  and  while  the 
current  decreases,  the  magnetic 
stored  energy  is  returned  to  the 
electric  circuit.  This  effect  of  the 
electric  current  on  the  surrounding 
space  is  termed  magnetic  induction. 
Unit  of  Inductance — When  a 
rate  of  change  of  current  of  one 
ampere  per  second  produces  an 
e.m.f.  of  one  volt,  the  circuit  is  said 
to  have  a  unit  of  inductance  called 
a  henry.     The  henry  being  incon- 

*Rcfercnccs: — For  a  bihliography  on  the  subject  of  skin 
effect  see  article  "Experimental  Researches  on  Skin  Effect  in 
Conductors"  by  A.  E.  Kennellv,  F.  A.  Laws,  and  P.  H.  Pierce 
in  A.  J.  E.  E.  Trans.,  Vol.  34,  Part  II  of  Sept.  1915.  .This 
article  ends  with  a  bibliography  on  the  subject  embracing  a 
very  complete  list  of  articles. 

"Calculation  of  Skin  Effect  in  Strap  Conductors"  by  H. 
B.  Dwight  in  Electrical  World,  March  II,  1916. 

"Skin  Effect  in  Tubular  and  Flat  Conductors"  by  H.  B. 
Dwight  m  A.  I.  E.  E.  Trans,  for  1918. 


For  various  sizes  of  solid  copper  rods.     For  stranded  conductors  of  equivalent  cross 
sectional  area  the  skin  effect  is  practically  the  same  as  for  the  solid  conductor. 


1! 

0 

hi 

=  S  " 

*«        0 

.5 

h 

si 

P 

J'ert-ent  Increase  of  Copper  Wires  Above  the  Direct- Current 

Resistance  Due  to  Alternating -Currents  of 

Different   Frequencies 

Hased   Upon   Direct  Current  Re- 
sistance at  25  Degrees  0. 
(77   Degrees  P.) 

Baaed  Upon  Direct  Current   Re- 
sistance at  65  Degrees  C. 
(149   Degrees   F.) 

V 

u 

si 

U 

U 

>> 

U 

(A 
U 

Si 

u 

4 

4 

n 

u 

■2  000  000 
1  800  000 
1  600  000 

1.631 
1.548 
1.459 

1.414 
1.342 
1.265 

2.2 
1.8 
1.4 

6.0 
4.9 
3.9 

14.1 

11.7 

9.4 

28.0 
23.7 
19.4 

78.6 
70.4 
61.4 

1.7 
1.3 
1.1 

4.S 
3.7 
3.0 

10.9 
9.0 
7.8 

22.1 
18.5 
15.0 

67.0 
60.0 
51.8 

1  500  000 
1  200  000 
1  000  000 

1.412 
1.263 
1.152 

1.225 
1.096 
1.000 

1.3 
0.8 
0.6 

3.4 
2.1 
1.5 

8.4 
5.5 
3.8 

17.4 
11.7 

8.4 

67.3 
42.7 
33.8 

0.9 
0.6 
0.4 

2.6 
1.7 
1.1 

6.4 
4.1 
3.0 

13.5 
9.0 
6.4 

47.4 
34.8 
26.2 

•.  50  000 
500  000 
250  000 

0.998 
0.815 
0.575 

0.866 
0.707 
0.500 

0.3 
0.1 

0.0 

0.9 
0.4 
0.1 

2.2 
1.0 
0.3 

4.9 
2.2 
0.6 

20.6 

10.1 

2.7 

0.2 
0.1 
0.0 

0.7 
0.8 
0.1 

1.7         3.7 
0.7         1.7 
0.2         0.4 

16.4 
7.7 
2.0 

curve  rneasured  vertically  the  intensity  of  the  field  at  the 
corresponding  distance.  The  radius  of  the  conductor 
has  been  assumed  as  unity,  and  maximum  field  density 
(always  at  the  surface  of  the  conductor)  as  100  percent. 
The  intensity  of  the  magnetic  field  starts  at  zero  at 
the  conductor  center,  and  increases  (with  uniform  dis- 
tribution of  current  in  the  conductor)  directly  as  the 


4  RESISTANCE—SKIN  EFFECT— INDUCTANCE 

distance   from  its  center  until   its   surface  is   reached,  The  intensity  of  the  magnetic  field  at  any  point  is 

where  it  becomes  maximum.     For  distances  beyond  the  proportional  to  the  m.m.f.  acting  at  that  point  and  in- 

surface  of  the  conductor,  the  field  intensity  varies  in-  versely  proportional  to  the  length  of  its  circular  path 

versely  as  the  distance  from  its  center.  (magnetic    reluctance).     Thus   at   the    surface   of    the 

TABLE  I— RESISTANCE  PER  1000  FEET 

OF    COPPER   CONDUCTORS   AT   VARIOUS   TEMPERATURES 

STRANDED   CONDUCTORS 


d 

z 

CO 

'^ 

00 

AREA 

CIRCULAR 
MILS 

OHMS   PER  1000   FEET  OF  SINGLE  CONDUCTOR 

ANNEALED  COPPER 

100%  CONDUCTIVITY 

HARD  DRAWN  COPPER 

97.3^  CONDUCTIVITY 

CO 

32°  F 

I5°C 

59°  F 

20''C 

68°F 

25°C 

77°  F 

35°C 

95°F 

50°  C 

I22°F 

65°C 

149°  F 

75°C 
I67°F 

0°C 
32°  F 

I5°C 

59°  F 

20°C 
68°  F 

25°C 
77°  F 

35°C 
95°F 

50°C 

I22°F 

65°C 
I49°F 

75°0 
I67°F 

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00  880 

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0101 

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1  loo  ooo 

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232 
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These  resistance  values  do  not  take  into  account  skin  effect.  This  should  be  considered  when  the  larger  conductors  are  used,  particu- 
larly at  the  higher  frequencies.  No  allowance  has  been  made  for  increased  length  due  to  sag  when  the  conductors  are  suspended..  The  resist- 
ance values  for  the  stranded  conductors  are  two  percent  greater  than  for  a  solid  rod  of  cross -section  equal  to  the  total  cross -section  of 
the  wires  of  the  cable.  ^ 

The  change  of  resistivity  of  copper  per  degree  C.  is  a  constant  independent  of  the  temperature  of  reference  and  of  the  sample  of  cop- 
per. This  resistivity -temperature  constant  is  0.0409  ohm  (mil,  foot).  The  fundamental  resistivity  used  in  calculating  this  table  is  the  an- 
nealed  copper   standard,   viz.   0.15328   ohm    (meter,    gram)    at   '20   degrees  C. 

For   sizes   not   given   in   the   table   computations  may  be  made  by  the  following  formulas  which  were  used  in  calculating  the  above  table:  — 


Ohms  per  1000  feet  of  annealed  copper  at  25  degrees  C  =- 


10787 


Circ.   mils 


at  65  degrees  C  — 


12457 


Circ.  mili 


RESISTANCE—SKIN  EFFECT— INDUCTANCE 


conductor  the  m.m.f.  reaches  its  maximum  because  all 
of  the  current  of  the  conductor  is  acting  to  produi^e 
m.m.f.  at  this  and  all  points  beyond.  On  the  other 
hand  the  circular  path  subject  to  this  maximum  m.m.f. 
is  shortest  at  the  surface,  the  reluctance  a  minimum 


and  consequently  the  field  intensity  is  greatest.  For 
points  beyond  the  surface  the  length  of  the  circular 
path  through  air  is  proportional  to  the  distance  from 
the  center  of  the  conductor.  Thus  at  a  distance  of  ^ 
from  the  center  the  circular  path  is  twice  as  long  as  at 


TABLE  II— RESISTANCE  PER  MILE 

OF    COPPER    CONDUCTORS    AT    VARIOUS    TEMPERATURES 

STRANDED    CONDUCTORS 


d 

z 

CO 

AREA 

CIRCULAR 
MILS 

OHMS   PER   MILE   OF  SINGLE  CONDUCTOR           1 

ANNEALED  COPPER 

100%    CONDUCTIVITY 

HARD  DRAWN  COPPER 

97.3%  CONDUCTIVITY 

0°C 
32°  F 

I5°C 
59°  F 

20°C 
68°F 

25°C 

77°  F 

35°C 
95°F 

50°C 
I22°F 

65°C 
I49°F 

75°C 
I67°F 

0°C 
32°F 

I5°C 
59°  F 

20°C 
68°F 

25°C 
77°  F 

35°C 

95°F 

50°C 
I22°F 

65''C 
I49°F 

75''0 
I67°F 

0000 

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1  cfoo  000 

1  800  000 

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These  resistance  values  do  not  take  into  account  skin  effect.  This  should  be  considered  when  the  larger  conductor,  are  used  partica- 
larly  at  the  higher  frequencies.  No  allowance  has  been  made  for  increased  length  due  to  sag  when  '*«  '^""''"'^'^^ " ^  »P«°''~-„..\''»  T"'": 
anc6  values  for  the  stranded  conductors  are  two  percent  greater  than  for  a  solid  rod  of  cross -section  equal  to  the  total  crosi  •  section  of 
the  wires  of  the  cable. 

The  change  of  resistivity  of  copper  per  degree  C.  is  a  constant  independent  of  the  temperature  »'  "'«"°", J»°^,"'  '»'VmT''u  th.'^." 
per.  This  resistivity  temperature  constant  is  0.0409  ohm  (mil.  foot).  The  fundamental  re.i.tivity  used  m  calculating  this  Ubl*  is  Us  u- 
nealed   copper   standard,  viz.   0.15328   ohm    (meter,   gram)    at  20   degrees  0. 


RESISTANCE— SKIN  EFFECT— INDUCTANCE 


a  distance  of  i  (its  surface)  and  consequently,  although 
the  m.m.f.  is  the  same  the  reluctance  is  double,  per- 
mitting only  one-half  as  great  a  flux  to  flow  as  at  the 
surface.  For  a  similar  reason  the  density  of  the  field 
at  a  distance  of  lo  is  one-tenth  the  surface  density;  at 
50  it  is  one-fiftieth,  etc.  The  curve  of  field  density  be- 
yond the  surface  of  the  conductor  therefore  assumes 
the  form  of  a  hyperbola. 

Inside  conductor  A  the  field  density  is  represented 
by  a  straight  line  joining  the  center  of  the  conductor 
to  the  apex  of  the  density  curve,  represented  as  100  per- 
cent.    Suppose  it  is  desired  to  determine  the  field  den- 

CHART  l-INDUCTANCE 


The  m.m.f.  resulting  from  equal  currents  is  the 
same  for  all  sizes  of  conductors.  Thus  the  field  den- 
sity at  points  equally  distant  from  the  center  of  dif- 
lerent  sizes  of  conductors  carrying  equal  currents  is 
equal  provided  these  points  lie  beyond  the  surface  of 
the  larger  conductor.  For  points  equally  distant  from 
the  center  of  different  size  conductors  which  lie  inside 
the  conductors  the  density  will  be  different.  Thus  if 
the  conductor  diameter  carrying  equal  current  be  re- 
duced to  one  half,  the  m.m.f.  at  its  surface  will  remain 
the  same,  but  since  the  flux  path  at  the  surface  is  now 
only  one-half  as  long,  the  flux  density  at  the  surface 


rioo% 


S  « 


PHYSICAL  PICTURE  ILLUSTRATING  THE 
MAGNETIC  FLUX  DENSITY  SURROUND- 
ING A  CONDUCTOR  CARRYING  DIRECT 
CURRENT  (SKIN  EFFECT  ABSENT!  WITH 
EFFECT  OF   RETURN  CIRCUIT  IGNORED. 


DOTTED  AREA  REPRESENTS  THE  MAGNETIC  FLUX  FIELD 
RESULTING  FROM  CURRENT  IN  CONDUCTOR  A  -  THE 
PORTION  OF  THIS  AREA  CONTAINING  CROSS  SECTION 
LINES  REPRESENT  THE  FLUX  DUE  TO  CURRENT  IN  CON- 
DUCTOR A  WHICH  IS  EFFECTIVE  IN  PRODUCING  INDUCT- 
ANCE IN  CONDUCTOR  A.  THIS  EFFECTIVE  FLUX  AREA 
IS  DIVIDED  INTO  THREE  SECTIONS:  TO  THE  LEFT  THAT 
EFFECTIVE  WITHIN  CONDUCTOR  A.  IN  THE  MIDDLE  THAT 
EFFECTIVE  BETWEEN  THE  TWO  CONDUCTORS  AND  TO 
THE  RIGHT  THAT  WHICH  IS  EFFECTIVE  THRU  CON- 
DUCTOR B, 


FORMULA 

TOTAL  INDUCTANCE  iN  MILLIHENRIES  (L)  PER  1000  FEET  OF  EACH  CONDUCTOR   L-0.0I6S4- 

sity  at  a  point  midway  between  the  center  and  surface 
of  the  conductor.  At  this  point  the  length  of  the  cir- 
cular path  is  one  half  its  length  at  the  conductor  sur- 
face. Since  the  current  distributes  uniformly  through- 
out the  cross-section  of  the  conductor,  at  a  point  mid- 
way between  the  center  and  its  surface,  one-fourth  of 
the  total  current  would  be  embraced  by  the  circle.  The 
m.m.f.  corresponding  to  this  point  would  therefore  be 
one-fourth  its  value  at  the  surface.  With  one-fourth 
m.m.f.  and  one-half  the  surface  reluctance  the  result- 
ing density  will  be  one-half  of  its  surface  density  as 
shown  by  this  value  falling  on  the  straight  line  at  this 
distance  from  the  center. 


3.1403  LOG      0 
10   R 


FIG.      6 


will  be  twice  as  great.  In  other  words,  the  magnetic 
field  density  at  the  surface  of  conductors  having  differ- 
ent diameters  but  carrying  the  same  currents  is  in- 
versely proportional  to  their  radii. 

The  area  indicated  by  cross-sectional  lines  on  the 
inductance  chart  represents  the  amount  of  inductance 
effective  in  conductor  A  resulting  from  current  in  this 
conductor.  It  will  be  seen  that  the  total  area  between 
the  adjacent  surfaces  of  the  conductors  /  to  9  below 
the  flux  density  line  is  effective.  This  part  of  the  in- 
ductance follows  a  logarithmic  curve  as  illustrated  on 
the  chart  and  is  represented  by  the  formula. 
D-R 
'~R'  


L  =  0.1403J  logn 


.(/) 


RESISTANCE—SKIN  EFFECT— INDUCTANCE 


Where  L  is  in  millihenries  per  looo  feet  of  single 
conductor. 

The  effective  flux  area  departs  from  the  flux  den- 
sity line  at  E  dropping  down  in  the  form  of  a  reverse 
curve  and  terminating  in  zero  at  //.  All  flux  to  the 
right  of  //  cuts  the  whole  of  both  conductors  producing 
the  same  amount  of  inductance  in  both  of  them  in  such 
a  direction  as  to  oppose  or  neutralize  each  other. 

The  flux  cutting  conductor  B  from  p  to  ii  has  its 
full  value  of  effectiveness  in  producing  inductance  in 
conductor  A.  On  the  other  hand  it  also  produces  to 
a  less  extent  inductance  in  conductor  B  but  in  a  direc- 
tion to  oppose  that  which  it  produces  in  conductor  A. 
The  difference  between  that  produced  in  conductors  A 
and  B  is  the  effective  flux  producmg  inductance  in  the 
circuit  and  is  represented  by  the  shaded  portion  through 
conductor  B  within  the  area  E-p-ii-T-E.  To  illustrate 
how  the  effective  flux  curved  line  E-T-ji  was  deter- 
mined, suppose  it  is  required  to  determine  the  effective 
flux  at  the  distance  lo  (center  of  conductor  B).  At 
this  point  the  flux  density  is  ten  percent,  but  since 
these  flux  density  lines  are  actually  concentric  circles, 
having  their  center  at  the  middle  of  conductor  A  this 
flux  density  curve  cuts  conductor  B  in  the  form  of  an 
arc  (see  lower  right  hand  corner  of  inductance  chart). 
The  area  of  the  shaded  portion  between  the  two  arcs  is 
a  measure  of  the  inductance  in  conductor  B  at  its  cen- 
ter. The  difference  between  this  shaded  area,  and  the 
whole  area  of  B,  or  the  clear  part  to  the  right  of  the 
shaded  portion,  is  a  measure  of  the  difference  in  induct- 
ance of  the  two  conductors.  In  other  words,  for  the 
spacings  shown,  approximately  55  percent  of  ten  or 
5.5  percent  is  the  value  of  the  effective  flux  at  distance 
of  10  from  conductor  A. 

,r  ■      ,         ,  ,  ,  ^-■^ 

If  in  place  of  L  =  o./^oj^  logw  =  — ^    (/) 


we  take  /,  =  o.r^o^y  logm        —^ (^) 

we  include  all  of  the  inductance  area  out  to  the  vertical 
Hne  O-io.  This  would  include  the  area  E-O-T  but  not 
the  area  T-io-ii.  Since  these  two  areas  are  equal,  the 
omission  of  one  is  balanced  by  including  the  other  and 
therefore  formula  (2)  correctly  takes  into  account  all 
of  the  effective  inductance  beyond  the  surface  of  con- 
ductor A. 

The  inductance  within  conductor  A  is  determined 
as  follows : — At  a  point  midway  between  the  center  and 
its  surface  the  flux  density  is  50  percent  as  indicated  by 
the  straight  flux  density  line  of  the  chart.  However  at 
this  point  only  one-fourth  of  the  conductor  area  is  en- 
closed, so  that,  measured  in  terms  of  its  effect  if  outside 
the  conductor,  its  effectiveness  would  be  only  one-fourth 
of  50  or  12.5  percent.  This  is  the  reason  that  the  so- 
called  effective  flux  line  is  curved  and  falls  to  the  right 
of  the  straight  flux  density  line.  The  area  of  the  trian- 
gular section  O-i-ioo  is  a  measure  of  the  effective  in- 
ductance within  conductor  A.  This  is  a  constant  for 
all  sizes  of  solid  conductors  and  is  represented  by  the 


constant  0.01524  of  the  inductance  formula  based  upon 
1000  feet  of  conductor. 

The  fundamental  formula  for  the  total  effective 
inductance  (within  and  external  to  conductor  A)  ol  a. 
single  solid  non-magnetic  conductor  suspended  in  air  >5 
therefore : 


/.  =  ox)i524  +  0.14037  logm  -j^periooo/l. 
or 

D 


(J) 


L  =  0.08047  +  0.7411S  logw-/^  per  mile (^) 

It  may  be  interesting  to  note  here  that  the  above 
described  graphical  solution  for  inductance  produces  re- 
sults in  close  agreement  with  these  obtained  by  the  fun- 
damental formula  for  inductance.  That  is,  lay  out  such 
a  chart  on  cross  section  paper  to  a  large  scale  and  count 
the  number  of  squares  or  area  representing  the  internal 
and  the  external  inductance  due  to  current  in  conductor 
A.  It  will  be  seen  that  the  relative  values  of  the  ex- 
ternal and  internal  flux  areas  conform  with  the  relative 
values  as  determined  by  the  formula.  This  will  also 
be  true  in  the  case  of  the  conductors  when  so  placed  as 
to  give  zero  separation,  as  illustrated  by  Fig.  6. 

VARIATIONS   FROM   THE   FUNDAMENTAL   INDUCTANCE 
FORMULA 

It  has  been  proven  mathematically  by  the  Bureau 
of  Standards  and  others  that  the  fundamental  formula 
(3)  for  determining  inductance  will  give  e.xact  results 
for  solid,  round,  straight,  parallel  conductors,  provided 
skin  and  proximity  effects  are  absent.  Proximity  ef- 
fect is  the  crowding  of  the  current  to  one  side  of  a 
conductor,  due  to  the  proximity  of  another  current 
carrying  conductor.  It  is  similar  to  skin  effect  in  that 
it  increases  the  resistance  and  decreases  the  inductance. 
Proximity  effect  as  well  as  skin  effect  changes  only  the 
inductance  due  to  the  flux  inside  the  conductor.  Proxi- 
mity effect  is  more  pronounced  for  large  conductors, 
high  frequencies  and  close  proximity. 

For  No.  0000  solid  conductors  at  zero  separation 
and  60  cycles,  the  error  in  the  results  (as  determined  by 
the  fundamental  inductance  formula)  due  to  skin  ef- 
fect is  less  than  one-tenth  of  one  percent.  This  error, 
however,  increases  rapidly  as  the  size  of  the  conductor 
increases.  Proximity  effect  cannot  be  calculated  but  it 
is  believed  to  be  less  than  two  percent  in  the  above  case. 

Should  skin  and  proximity  effect  combined,  be 
sufficient  to  force  all  of  the  current  out  to  within  a  verv 
thin  annulus  at  the  surface  of  the  conductor  (a  condi- 
tion obviously  never  obtained  at  commercial  frequen- 
cies) their  combined  effect  would  be  a  maximum.  In 
such  a  case  there  would  be  no  inductance  within  the 
conductors  and  the  first  constant  0.01524  would  dis- 
appear from  formula  (3). 

Skin  and  proximity  effect  are  so  small  in  the  case 
of  the  greater  spacings  of  conductors  required  for  high- 
tension  aerial  transmission  circuits  that  they  may  in 
such  cases  be  ignored.     Even  in  the  case  of  the  close 


RESISTANCE—SKIN  EFFECT— INDUCTANCE 


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RESISTANCE— SKIN  EFFECT— INDUCTANCE 


spacings  required  for  three  conductor  cables  these  com- 
bined effects  are  usually  less  than  four  percent. 

EFFECT   OF   STRANDING   ON   INDUCTANCE 

The  fundamental  formula  (3)  for  determining  in- 
ductance is  based  upon  a  solid  conductor,  R  being  taken 
as  the  radius  of  the  conductor.  In  stranded  cables  the 
effective  value  for  R  lies  between  the  actual  radius  and 
that  of  a  solid  rod  having  an  equivalent  cross-section  10 
that  of  the  cable.  The  effective  value  for  R  varies  with 
the  stranding  of  the  cable  employed. 

Formulas  for  use  in  determining  the  inductance  of 
stranded  cables  when  used  for  high-tension  aerial  trans- 
mission have  been  calculated  by  Mr.  H.  B.  Dwight  as 
follows : — 

For  a    7-wire  cable,  L  =  0.741  logjo  -^-^ — (5) 

_  .  , .      ,  2.640  D 

For  a  19-wire  cable,  L  —  0.741  logw -3 {6) 

T.  ■  r  ,  2-605    D 

For  a  37-wire  cable,  L  =  0.J41  logw  — ^ (7) 

2590  D 
For  a  6i-wire  cable,  L  =  o.jtr  logya  — j {S) 

where  L  is  in  millihenries  per  mile  of  a  single  conductor, 
D  is  the  spacing  between  centers  of  cables,  and  d  is  the 
outside  diameter  of  the  cables  measured  in  same  units 
as  D. 

SPIRALING   EFFECT   UPON    INDUCTANCE 

Spiraling  of  the  strands  of  a  cable  and  spiraling  of 
the  conductors  of  a  three-conductor  cable  tend  to  in- 
crease the  inductance.  It  is  difficult  to  calculate  the 
viffect  of  spiraling  for  the  various  cases,  but  it  may  be 
considered  negligible  for  high-tension  aerial  transmis- 
sion circuits  using  non-magnetic  conductors.  For 
three-conductor  cables  the  effect  of  spiraling  is  probably 
in  the  neighborhood  of  two  percent. 

Values  for  inductance  per  thousand  feet  of  single 
conductor  are  given  in  Table  III,  for  commercial  sizes 
of  copper  and  steel  reinforced  aluminum  conductors. 
The  formula  by  which  the  values  were  derived  are: — 

L  =  oj3r524  -f-  0.1403  logv)  -^ 0) 

where  L  =  Millihenries  per  1000  feet  of  single  conductor  of  a 
single  phase,  or  of  a  symetrical  three-phase  circuit. 

D  =  Distance  between  centers  of  conductors. 

R  =  Radius  (to  be  measured  in  same  units  as  D)  of 
solid  conductor.  In  the  case  of  stranded  con- 
ductors, A'  was  taken  as  the  radius  of  a  solid  rod 
of  equivalent  cross-section  to  that  of  the  stranded 
conductors. 

Table  III  has  been  carried  out  to  three  figures  only. 
This  would  seem  sufificiently  accurate  for  working 
values  when  it  is  considered  that  there  are  numerous 
sources  of  variation  from  the  calculated  values.  In 
the  first  place  formulas  are  based  upon  a  uniform  dis- 
tribution of  current  throughout  the  cross  section  of  the 
•onductors,  whereas  the  current  is  seldom  uniform  and 
n  the  larger  conductors,  especially  at  60  cycles,  may  be 
to  a  large  extent  crowded  to  the  outer  strands  as  a 
result  of  skin  effect.     This  condition  is  further  modi- 


fied when  the  conductors  are  placed  close  together,  by 
the  pro.ximity  effect.  Stranded  conductors  made  up  of 
various  stranding  combinations  result  in  variation  of 
inductance  of  several  percent.  In  practice  the  length 
and  spacing  of  conductors  will  vary  more  or  less  from 
those  assumed  when  determining  the  calculated  values. 

The  values  for  inductance  of  stranded  conductors 
in  Table  III,  as  stated  above,  were  derived  by  taking  R 
as  the  radius  of  a  solid  rod  having  an  equivalent  cross- 
section  area  to  that  of  the  stranded  conductors.  Thus 
for  1 000  000  circ.  mil  cable  the  outside  diameter  is 
1. 152  in.  and  that  of  an  equivalent  solid  iron  is  i.o  in. 
R  was  therefore  in  this  case  taken  as  0.5  in.  The  ef- 
fective radius  is  really  slightly  greater  than  that  of  the 
solid  rod  and  less  than  that  of  the  cable,  varying  with 
the  stranding  employed.  The  actual  inductance  of 
cables  will  therefore  be  slightly  less  (usually  two  or 
three  percent)  than  those  indicated  in  the  table  for 
solid  rods.     The  table  values  are  therefore  conservative. 

The  steel  core  of  steel  reinforced  aluminum  cables 
carries  so  little  current  on  account  of  its  relatively 
greater  resistance  that  for  practical  purposes  it  has  been 
customary  to  ignore  its  presence  and  to  consider  such 
conductors  as  solid  rods  of  same  area  as  that  of  the 
aluminum  strands.  In  the  absence  of  accurate  data 
this  practice  was  followed  in  determining  the  values  for 
inductance  of  such  cables  in  Table  III. 

The  minimum  value  for  inductance  occurs  when 

the  conductors  have  zero  separation -„■  =  z,  (Fig.  6). 
In  this  case  the  inductance  in  millihenries  is  independent 
of  the  size  of  the  conductor.  As  given  by  formula  (3) 
it  is  L  =  0.05124  -\-  O.J403  log^a  2  =  0.0575  millihen- 
ries per  1000  feet  of  each  conductor.  Obviously  in- 
sulation requirements  will  not  permit  of  such  a  low 
value  for  inductance  although  it  will  be  closely  ap- 
proached in  low  voltage  cables. 

Any  given  percentage  difference  in  distance  be- 
tween centers  of  conductors  represents  a  definite  and 
constant  value  in  inductance  regardless  of  their  siz*?. 
These  values  are  given  in  column  B  at  the  bottom  of  the 
table  for  various  percentages  increase  in  spacings. 
Thus  if  the  distance  between  conductor  centers  is  in- 
creased 50  percent  the  corresponding  increase  in  in- 
ductance is  0.025  as  indicated  in  column  B,  under  the  A 
values  of  1.50.  Likewise  doubling  the  distance  in- 
creases the  inductance  by  an  amount  of  0.042.  For  in- 
stance the  table  value  for  inductance  of  No.  o  solid 
copper  conductor  is  for  one-half  inch  spacing  0.084,  and 
for  one  inch  spacing  0.126  (an  increa.se  of  0.042.)  For 
four  foot  spacing  the  table  value  is  0.362,  and  for  eight 
foot  spacing  0.404,  also  a  difference  of  0.042. 

References: — An  article  by  Prof.  Giarles  F.  Scott,  "In- 
ductance in  Transmission  Circuits"  in  The  Electric  Journ.\l 
for  Feb.  1906  very  clearly  covers  the  field  of  self  and  mutual 
inductance  external  to  the  conductors. 

H.  B.  Dwight,  "Transmission  Line  Formulas." 
V.  Karapetoff,  "The  Magnetic  Circuit"  p.   189. 


CHAPTER  II 
REACTANCE-CAPACITANCE-CHARGING  CURRENT 


REACTANCE 


A  CONDUCTOR  carrying  an  electric  current  is 
surrounded  by  a  magnetic  flux,  whose  value  is 
proportional  to  the  current.  If  the  current 
varies,  this  flux  also  changes,  thereby  inducing  an  elec- 
tromotive force  in  a  direction  which  opposes  the  change. 
This  counter  e.m.f.  is  proportional  to  the  rate  of  change 
and  hence  in  alternating  current  is  proportional  to  the 
frequency.  It  can  be  expressed  in  ohms  per  mile  of 
each  conductor  of  a  single-phase  or  of  a  symmetrical 
three-phase  circuit  as  follows: — 

Ohms  Reactance  =  ^  tt  /  L  (9) 

When  /  =  Frequency  in  cycles  per  second 

L  =  Henries  per  mile  of  single  conductor. 

The  value  for  2    -n-  f  are  as  follows : — 

Frequency  2  v  f 

I  6.28 

15  9425 

25  157.1 

40  251.3 

60  3770 

133  835-7 

Tables  IV  and  V  indicate  the  reactance  in  ohms 
per  mile,  of  a  single  conductor  at  25  and  60  cycles  re- 
spectively for  various  spacings  of  conductors.  The 
foot  notes  to  these  tables  cover  the  pertinent  points  re- 
lating to  them.  The  resistance  per  1000  feet,  and  per 
mile  of  single  conductor  at  25  degrees  C.  {"jj  degrees 
F)  is  given  in  parallel  columns  as  a  convenience  for 
comparison  of  the  resistance  and  reactance  values.  The 
resistance  corresponding  to  other  temperatures  when 
desired  may  be  taken  from  Tables  I  and  II. 

Tables  VI  and  VII  indicate  the  relative  importance 
of  reactance  and  resistance.  In  some  cases  of  short 
lines  and  large  single  conductors,  the  reactance  and  not 
the  resistance  may  determine  the  size  and  number  of 
cables  necessary.  In  other  words,  it  may  be  necessary 
to  keep  the  resistance  abnormally  low  so  that  the  react- 
ance will  not  be  so  high  as  to  result  in  an  abnormal  volt- 
age drop  in  the  circuit.  In  such  cases  the  values  in 
Tables  VI  and  VII  may  be  used  for  determining  the 
permissible  resistance  in  order  not  to  exceed  the  desired 
reactance. 

Example: — It  is  desired  to  use  1000 000  circ.  mil  single 
conductor  cables  at  60  cycles,  spaced  two  feet  apart;  from 
Table  VII  it  is  seen  that  the  reactance  drop  under  these  condi- 
tions is  8.52  times  the  ohmic  drop  at  25  degrees  C.  If  an  ohmic 
drop  of  five  percent  at  25  degrees  C  is  suggested  the  corre- 
sponding reactive  drop  would  be  5  X  8.52  or  42.6  percent  which 
would  be  excessive.  If  it  is  desired  to  limit  the  reactive  drop 
to  ID  percent  in  this  case,  the  ohmic  drop  at  25  degrees  C  must 
be  10  H-  8.52  or  1. 18  percent. 

Probably  a  more  important  use  for  Tables  VI  and 
VII  is  for  determining  the  reactance  of  a  conductor  di- 
rectly from  its  resistance.  To  do  this  it  is  only  neces- 
sary to  multiply  its  resistance  (at  25  degrees  C)  by  the 


ratio  value  in  table  VI  or  VII  corresponding  to  the  con- 
ductor and  spacing  desired. 

UNSYMMETRICAL  SPACING 

The  inductance  and  capacitance  per  conductor  of 
a  three-phase  circuit  for  symmetrical  spacing  of  con- 
ductors is  the  same  as  the  inductance  and  capacitance 
per  conductor  of  a  single-phase  circuit  for  the  same  size 
conductor  and  the  same  spacing.  For  irregular  spac- 
ing of  conductors,  the  inductance  and  capacitance  will 
be  different.  When  the  three  conductors  are  placed  in 
the  same  plane  (flat  spacing),  the  inductance  of  each  of 
the  outside  conductors  is  greater  than  that  of  the  middle 
conductor.  By  properly  transposing  the  conductors, 
the  inductance  and  capacitance  may  be  equalized  in  all 
three  conductors.     However,  the  effect  of  flat  spacing 

^— A,--Y- e -» 

9 • 1 


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1 

Conductor  Spacings. 

For  three-phase  irregular  flat  or  triangular  spacing  (Figs. 
7  and  8)  use  D  =  ^' A  B  C. 

For  three-phase  regular  flat  spacing  Fig.  9  use  D  ^  1.26  A. 

For  two-phase  line  the  spacing  is  the  average  distance  be- 
tween centers  of  conductors  of  the  same  phase.  It  makes  no 
difference  whether  the  plane  of  the  conductors  with  flat  spacing 
is  horizontal,  vertical  or  inclined. 

is  equivalent  to  that  of  a  symmetrical  arrangement  of 
greater  spacing. 

Various  arrangements  of  conductors  are  indicated 
in  Figs.  7,  8  and  9.  Many  three-phase  high  tension 
circuits  have  the  three  conductors  regularly  spaced  in 
a  common  plane  (regular  flat  spacing)  Fig.  9.  Beneath 
these  figures  are  placed  statements  indicating  the  de- 
termination of  "effective  spacings"  for  any  arrange- 
ment of  conductors. 

Since  the  so  called  "effective  spacing"  correspond- 
ing to  unsymmetrical  arrangements  of  conductors  is 
usually  a  fractional  number,  the  line  constants  for  such 
effective  spacing  can  usually  not  be  taken  directly  from 


REACTANCE— CAPACITANCE— CHARGING  CURRENT 


II 


the  tables  but  may  be  obtained  by  the  use  of  the  values 
in  columns  A  and  B  at  the  foot  of  these  tables. 

Example : — It  is  desired  to  determine  the  60  cycle  reactance 
per  mile  of  a  single  conductor  for  flat  spacing  of  11  ft.  between 
adjacent  0000  solid  copper  conductors.  The  effective  spacing 
is  1.26  X  II  or  13.8  feet.  The  reactance  (Table  V)  for  this 
conductor  at  13  feet  symmetrical  spacing  is  0.820  ohm.  The 
value  for  A,  (bottom  of  Table  V)  =  13.8  -^  13  =  1.06.  The 
value  of  B  corresponding  to  the  value  for  A  of  1.06  is  approxi- 
mately 0.006  which,  added  to  0.820  gives  a  reactance  of  0.826 
ohm  for  the  effective  spacing  of  13.8  feet.  The  values  of  re- 
actance for  all  effective  spacings  not  included  in  the  Table 
may  be  determined  in  a  similar  manner. 

With  an  unsymmetrical  arrangement  of  conductors 
there  must  be  a  sufficient  number  of  transpositions  of 
conductors  to  obtain  balanced  electrical  conditions  along 
the  circuit. 

CAPACITANCE 

When  mechanical  force  is  exerted  against  a  liquid 
or  a  solid  mass,  a  displacement  takes  place  proportional 
to  the  force  exerted  and  inversely  proportional  to  the 
resistance  offered  by  the  liquid  or  solid  mass  subjected 
to  the  force.  If  the  mass  consists  of  some  elastic  ma- 
terial, such  as  rubber,  the  displacement  will  be  greater 
than  if  it  consists  of  a  more  solid  material,  such  as 
metal. 

In  a  similar  manner  when  an  e.m.f.  is  applied  to  a 
condenser,  a  certain  quantity  of  electricity  will  flow  into 
it  until  it  is  charged  to  the  same  pressure  as  that  of  ihe 
applied  circuit.  A  condenser  consists  of  plates  of  con- 
ducting material  separated  by  insulating  material  known 
as  the  dielectric.  All  electric  circuits  consist  of  conduc- 
tors separated  by  a  dielectric  (usually  air)  and  there- 
fore act  to  a  greater  or  less  extent  as  condensers.  The 
ability  of  a  condenser  or  any  electric  circuit  to  receive 
the  charge  is  a  measure  of  its  "capacity"  more  properly 
known  as  its  "capacitance".  Just  as  the  rubber  mass 
referred  to  above  will,  for  a  given  force,  permit  of 
greater  displacement  so  will  circuits  of  greater  capaci- 
tance permit  more  current  to  flow  into  them  for  a  given 
e.m.f.  impressed. 

The  process  of  charging  a  dielectric  consists  of  set- 
ting up  an  electric  strain  in  it  similar  to  the  mechanical 
strain  in  a  liquid  or  mass  referred  to  above.  If  an  al- 
ternating voltage  is  impressed  upon  the  terminals  of  a 
circuit  containing  capacitance,  the  charging  current  will 
vary  directly  with  the  impressed  e.m.f.  There  is  cur- 
rent to  the  condenser  during  rising  and  from  the  con- 
denser during  decreasing  e.m.f.  Thus  the  condenser  is 
charged  and  then  discharged  in  the  opposite  direction 
during  the  next  alternation,  making  two  complete 
charges  and  discharges  for  each  cycle  of  impressed 
e.m.f.  (Fig.  10).  As  long  as  the  e.m.f.  at  the  terminals 
is  changing,  the  condenser  will  continue  to  receive  or 
give  out  current.  The  current  flowing  to  and  from  the 
condenser,  assuming  negligible  resistance,  leads  the  im- 
pressed e.m.f.  by  90  electrical  degrees. 

DEFINIITION 

The  capacitance  of  a  circuit  or  condenser  is  said  to 
be  one  farad  when  a  rate  of  change  in  pressure  of  one 
volt  per  second  at  the  terminals  produces  a  current  of 


one  ampere.  Stated  another  way,  its  capacitance  in 
farads  is  numerically  equal  to  the  quantity  of  electricity 
in  coulombs  which  it  will  hold  under  a  pressure  of  one 
volt.  The  farad  being  an  inconveniently  large  unit,  one 
millionth  part  of  it,  the  microfarad,  is  the  usual  prac- 
tical unit. 

CAPACITANCE  TOnUULA 

An  exact  formula  for  the  capacitance  between 
parallel  conductors  must  take  into  account  the  nonuni- 
formity  of  the  distribution  of  charge  around  the  con- 
ductors. Such  a  formula*  is  formed  by  considering  the 
charges  as  concentrated  at  the  inverse  points  of  the 
conductors;  thus, — 
0.00S46/ 


C  = 


COih'^-J 


(10) 


Where  C  equals  the  microfarads  per  1000  feet  of 
conductor  between  two  parallel  bare  conductors  in  air, 
D,  the  distance  between  centers  of  the  conductors  and 
d,  the  diameter  and  R  the  radius  of  the  conductors, 
measured  in  the  same  units  as  D. 

Since  Cosk-^ X  =  loge  {X -f-  1 '  A'--/) (//) 

0.008462 


C  = '- 


(") 


If— CHAROe- 
FIG.    10 — CH.^HGINC   CURBENT 


-CNSOHARQEh' 


Reducing  to  common  logarithms  and  capitancc  to 
neutral, — 

0.00735* 


(/i) 


Microfarads  'per  1000  feet  of  single  conductor  to  neutral. 


or 


C  = 


o.o^SSiC) 


'■"(4+v(^)"-') 


U4) 


Microfarads  per  mile  of  single  conductor  to  neutral. 

When  D  is  greater  than  10  d,  which  is  always  the 
case  in  high-tension  transmission  lines  employing  bare 
conductors,  the  following  simplified  formulas  may  be 
used  with  negligible  error. — 


0fiO7354 

D 

Ingvi  -^ 


as) 


♦See  article  by  Pender  &  Osborne  in  Electrical  World  of 
Sept.  22,  1910,  Vol.  56. 


REACTANCE— CAPACITANCE— CHARGING   CURRENT 


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SUSCEPTANCEiVALUES  OF  THE  SPACINGS  COMPARED.        THE  CHARGING  K.V.A.  3  PHASE  -  (CHARGING  CURRENT  IN    AMPEF 

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REACTANCE— CAPACITANCE— CHARGING   CURRENT 


Microfarads  per  icxx)  feet  of  single  conductor  to  neutral, 
or 

0.03SSJ 


r  = 


(i<5) 


logK    -jr 
Microfarads  per  mile  of  single  conductor  to  neutral. 

The  above  formulas  are  only  applicable  to  ordinary 
overhead  circuits  when  the  distance  from  the  conductor 
to  other  conductors,  particularly  the  earth,  is  large 
compared  to  their  distance  apart.  However,  since  the 
effect  of  the  earth  is  usually  small  in  most  practical 
cases,  the  formulas  give  a  very  close  approximation  to 
the  actual  capacitance  of  overhead  circuits. 

The  values  of  capacitance  in  Table  VIII  were  de- 
rived by  using  formula  (13).  For  calculating  the 
capacitance  for  the  stranded  conductors,  the  actual 
overall  diameter  of  the  cable  was  taken.  This  intro- 
duces a  small  error  which  is  negligible  except  for  very 
close  spacings  not  used  in  high  tension  transmission 
lines  employing  bare  conductors. 

CHARGING  CURRENT 

RELATION  OF  CHARGING  CURRENTS  OF  SINGLE  AND 
THREE-PHASE  SYSTEMS 

The  diagrams  (Fig.  11)  may  assist  in  forming  a 
clear  understanding  of  the  relation  of  charging  current 

< <?■ 


0.281  AMPERE 
CHARGING 
CURRENT 


50  000 
VOLTS 


100  000 
VOLTS 


■I- - 

50  000 

VOLTS   r: 


— 7^ — '^ 
1 00  000 

VOLTS 

1 00  000 
VOLTS 


1 00  000 
VOLTS 


system  is  15.5  percent  greater  than  in  the  single-phase 
system,  and  the  resulting  charging  k.v.a.  is  just  double 
that  of  the  single-phase  system.  The  charge  on  any 
particular  conductor  is  in  phase  with  the  voltage  be- 
tween that  conductor  and  the  neutral  and  the  charging 
current  for  that  conductor  is  90  degrees  ahead  of  the 
voltage  drop  from  that  conductor  to  neutral. 

Grounding  of  the  neutral  point  of  a  system  has  no 
effect  upon  the  charging  current  when  the  system  is  in 
static  balance.  In  determining  the  total  charging  cur- 
rent to  be  supplied  by  a  given  generating  station,  it 
should  be  remembered  that  in  cases  of  duplicate  trans- 
mission circuits,  when  both  circuits  are  excited,  the 
charging  current  will  be  approximately  double  what  it 
would  be  if  only  one  of  the  circuits  were  in  use. 

Tables  IX  and  X  contain  values  for  capacitance 
susceptance  to  neutral  in  micromhos  per  mile  of  con- 
ductor. As  indicated,  the  charging  current  in  amperes 
per  mile  of  single  conductor  to  neutral  =  the  (suscep- 
tance from  table)  X  (volts  to  neutral)  X  io-*.Thus 
in  a  three-phase,  60  cycle,  100  000  volt,  (57  740  volts  to 
neutral),  symmetrical  circuit,  the  No.  0000  stranded 
conductors  being  arranged  at  the  corners  of  an  equi- 
lateral triangle  spaced  nine  feet  apart,  the  charging  cur- 
rent per  mile  would  be  determined  as  follows: — 

S.62  y.  57  740  X  /o-«  =  0.3245 

amperes  to  neutral 

or  0.3245  X  5774"  =  ^S.737 

k.v.a.  to  neutral 

^S.737  X  ^  =  5i5.^  K.v.a. 

total  three  phase 

Table  XI  is  an  extens- 
ion of  Tables  IX  and  X 
from    which     values 


T 


0.324  AMPEPE  CHARGING  CURRENT 


z 


0.32A  AMPERE  CHARGING  CURRENT 


in 


SINGLE  PHASE  CIRCUIT  THREE  PHASE  CIRCUIT 

FIG.    II — CHARGING  CURRENT  IN  SINGLE  AND  THREE-PHASE  CIRCUITS 


to  susceptance  for  single  and  three-phase  circuits.  In 
the  following  consideration  No.  0000  stranded  copper 
conductors  will  be  assumed  as  spaced  nine  feet  between 
any  two  conductors,  frequency  60  cycles,  vol'.age 
100  000  volts  between  conductors.  Voltage  to  neutral 
will  therefore  be,  for  single  phase  circuit,  50000  volts 
and  for  three-phase  circuit  57  740  volts.  Distance  of 
transmission  one  mile.  From  Table  VIII,  a  capaci- 
tance to  neutral  of  0.00282  microfarads  per  1000  feet  is 
obtained  which  is  equivalent  to  0.0149  microfarads  per 
conductor  to  neutral  for  this  one  mile  of  circuit.  The 
susceptance  will  therefore  be  as  follows: — 

Per  conductor  to  neutral  2  ir  i  Cn  ^  5.62  microhms 
Between  conductors  2  tt  f  Qj  =  2.81  microhms 

For  Single-Phase  Circuit— To  neutral  5.62  X  5° 
000  X  10°  '^=  0.281  amperes  or  between  conductors 
2.81  X  -too  000  X  10'  =  0.281  amperes  therefore  charg- 
ing k.v.a.  is  0.281  X  50000  X  2  =  28.1  k.v.a.  single 
phase  or  0.281  X  100  000  ^  28.1  k.v.a.  single  phase. 

For  a  Three-Phase  Circuit — To  neutral  5.62  X  57 
740  X  10*  =  0.324  amperes.  Therefore  charging  k.v.a. 
is  0.324  X  57  740  X  3  =  56.2  k.v.a.  three-phase. 

It  will  be  seen  from  the  above  that  the  charging 
current  per  conductor  in  the  three-phase  symmetrical 


k.v.a.,  three-phase  for 
charging  current  have 
been  calculated  for  certain  assumed  spacings  and  aver- 
age voltages.  In  the  case  cited  above  it  was  found  that 
the  charging  current  would  be  56.2  k.v.a.,  three-phase 
per  mile.  Table  XI  gives  this  value  directly  for  the 
conditions  specitied. 

CHARGING  CURRENT  AT  ZERO  LOAD 

The  term  charging  current  of  a  transmission  circuit 
refers  to  the  amount  of  current  which  flows  into  the 
circuit  at  the  supply  end  with  normal  voltage  held  at 
the  receiver  end  at  aero  load.  If  the  circuit  is  long,  its 
capacitance  will  be  high  and  therefore  the  voltage  at 
the  supply  end  may  be  considerably  less  than  at  the  re- 
ceiver end.  For  instance  a  60  cycle  circuit  300  miles 
long,  having  certain  constants  will,  with  100  000  volts 
maintained  at  the  receiver  end,  have  a  voltage  of  only 
80000  volts  at  the  supply  end  at  zero  load.  This  same 
circuit  will  at  full  load  and  lOO  000  volts  maintained  at 
the  receiver  end,  require  120000  volts  at  the  supply 
end.  It  is  evident  therefore  that,  since  the  charging 
current  varies  with  the  voltage,  if  the  circuit  has  much 
capacitance  the  voltage  along  the  circuit,  and  particu- 
larly near  the  supply  end,  will  vary  to  a  large  extent 


REACTANCE— CAPACITANCE— CHARGING  CURRENT 


31 


and  consequently  the  charging  current  of  the  circuit 
will  be  different  for  different  loads. 

In  case  of  the  300  mile  circuit  referred  to  above, 
the  charging  current  at  zero  load  will  be  very  much  less 
than  it  is  at  full  load,  because  the  average  voltage  at 
zero  load  is  less  than  the  average  voltage  at  full  load. 
At  zero  load  the  average  voltage  is  less  and  at  full  load 
it  is  greater  than  the  receiver  end  voltage. 

It  is  customary  to  calculate  the  total  charging  cur- 
rent for  the  circuit  by  multiplying  the  total  susceptance 
by  the  receiver  end  voltage.  This  would  be  correct  if 
the  voltage  throughout  the  length  of  the  circuit  were 
held  constant  and  of  the  same  value  as  at  the  receiver 
end.  This  condition  is  approximately  met  within 
commercial  lines  and  this  method  of  determining  the 

70,- 


susceptance  by  the  receiver  voltage.  For  a  circuit  300 
miles  long  the  error  in  charging  current  is  only  two 
percent  for  25  cycles  and  seven  percent  for  60  cycle  cir- 
cuits. The  error  in  charging  k.v.a.  is  four  percent  for 
25  cycle  and  32  percent  for  60  cycle  circuits. 

RELATION  OF  INDUCTANCE  TO  CAPACITANCE 

As  conductors  are  brought  closer  together,  the  in- 
ductance decreases  and  the  capacitance  increases. 
These  values  change  with  changes  in  spacings  between 
conductors  in  such  a  mannar  that  their  product  L  X  C" 
is  practically  a  constant  for  all  spacings  (except  very 
close  spacings  such  as  used  in  low-voltage  service  and 
lead-covered  cables)  and  for  all  sizes  of  conductors. 
If  there  were  no  losses  encountered  by  the  electric 

''     propagation     in     the 

conductors  themselves 

the  product  of  L  and 

C  would  be  a  constant 

55  5  for   all   spacings  and 

i  sizes  of  conductors. 

=">  I  In  Table  C  is  in- 

o  dicated    the    relation 

g  of    the  total    induct- 

*°g  ance  and  capacitance, 

j6  o  and      their      product, 

~  in     two     bare     par- 

''  8  allel  conductors  in  air 
< 

25  a  for  a  circuit  one  mile 

< 

=  long.     The  values  for 

2C  y 


^ 


SUPPLV  END 


DISTANCE  IN  MILES  FROM  SUPPLY  END 
12 — CHARGING   CURRENT   AT   ZERO    LOAD   FOR    VARIOUS    LENGTHS 


At  zero  load  the  voltage  (on  account  of  the  effect  of  capacitance)  decreases  as 
the  supply  end  of  the  circuit  is  approached.  The  charging  current  at  points  along 
the  circuit  decreases  directly  as  the  voltage.  If  the  charging  current  for  zero  load  is 
estiinated  by  the  approximate  method  based  upon  the  receiver  voltage  being  main- 
tained throughout  the  length  of  the  circuit  the  result  will  be  too  high.  The  error  will 
increase  as  the  length  of  the  circuit  is  increased;  it  will  also  increase  rapidly  as  the 
frequency  is  raised.  The  error  in  the  resulting  K.V.A.  required  to  charge  the  circuit 
will  therefore  increase  very  rapidly  with  an  increase  in  distance  or  frequency.  The 
curves  below  represent  an  approximation  of  this  error. 


total  charging  current  is  therefore  sufficiently  accurate 
for  most  practical  purposes. 

For  the  purpose  of  making  exact  calculation  of  the 
total  current  at  the  supply  end  of  long  circuits,  the 
charging  current  must  be  calculated  by  mathematical 
formulas  which  accurately  take  into  account  the  change 
in  voltage  along  the  circuit  at  zero  load.  This  will  be 
taken  up  in  a  later  article.  It  may  be  interesting  to 
note  approximately,  however,  how  the  charging  cur- 
rent and  charging  k.v.a.,  as  determined  by  the  above 
method,  varies  from  what  it  would  be  if  calculated  by 
the  rigorous  formula.  The  curves  in  Fig.  12  represent 
an  approximation  to  the  error  when  calculating  the 
charging  current  at  zero  load  by  multiplying  the  total 


L  are  in  millihenries 
and  for  C  in  micro- 
farads. Since  the 
formulas  by  which  L 
and  C  were  calculated 
account  for  the  flux 
within  the  conductors 
themselves,  the  prod- 
uct LC  is  not  a  con- 
stant, as  will  be  seen 
by  the  tabulated 
values,  although  for 
the  larger  spacings 
such  as  used  in  high- 


tension  transmission  the  product  is  nearly  a  constant. 
TABLE  C— PRODUCT  OF  (TOTAL)  L  AND  (TOTAL)  C 


Solid  Conductors 

a" 

Induct- 
ance /. 
F'ormula 

(4) 

C.apac  tanc« 

Size 

Diam. 
Inches 

Formula 

Product 

1000  000 

I  000  000 

I  000  000 

0000 

uooo 

0000 

1. 00 
1. 00 
1. 00 
0.46 
0.46 
0.46 

2 

24 
300 

2 

24 
300 

1053 
2.653 
4.279 

1-553 
3153 
4-779 

0.0339S 
0.01 1 55 
0.00695 
0.02079 
0.00961 
0.00623 

0.03S7S 
0.03064 

0.02974 
0.03228 
0.03030 
a02977 

RELATION    OF    INDUCTANCE    AND    CAPACITANCE   TO   LIGHT 
VELOCITY 

The  propagation  of  the  electric  and  the  magnetic 


22 


REACT  A  NCE~CA  PA  CI  TA  .V  CE—CHA  RGING    C  URREN  T 


fields  in  a  dielectric,  such  as  air,  is  the  same  as  that  of 
light.  Along  a  transmission  line  it  is  retarded  only 
slightly  due  to  losses  or  the  fact  that  the  current  is  not 
confined  to  the  surface  of  the  conductors.  If  the  induc- 
tance inside  the  conductors  is  negligible,  then  the  ve- 
locity of  the  electric  and  the  magnetic  fields  is  the  same 
as  light,  that  is  approximately  i86  ooo  miles  per  second 
or  approximately  3  X  io^°  cm.  per  second.  For  high- 
tension  transmission  lines  of  large  spacings,  the  induc- 
tance inside  the  conductor  is  relatively  small,  so  that  the 
speed  of  the  electric  field  is  practically  that  of  light. 

The  following  relation  exists  between  inductance  L 
in  henries,  capacity  C  in  farads  and  velocity  of  light  V 
per  second: — 


LC  (in  air)    =  -jTT.or,  F  = 


]'  LC 


(17) 


Thus  it  will  be  seen  that  if  either  L  or  C  is  known, 
the  other  may  be  determined  since  the  velocity  of  light 
V  is  known.  If  values  for  L  and  C  are  taken  which  in- 
clude the  inductance  inside  the  conductors,  particularly 
if  the  conductors  are  very  close  together,  it  would  be 
necessary  to  assume  a  velocity  of  electric  propagation 


somewhat  less  than  that  of  light.  If,  on  the  other 
hand,  the  values  for  L  and  C  external  to  the  conductors 
are  taken,  then  the  above  equation  is  rigidly  correct. 

In  Table  C,  it  was  shown  that  for  No.  0000  con- 
ductors, 300  inch  spacing,  the  total  values  of  L  and  C 
were  for  a  single-phase  line, — 

L  =  0.004779  henries  per  mile  of  circuit. 
C  =  0.000  000  006  23  farads  per  mile  of  cir- 
cuit. 

therefore,    1    =  ..  ,       = 

y  o.oo^j/g  X  0.000 000 006 2 J 

183  000  miles  per  second (18) 

which  is  less  than  the  speed  of  light. 

If  we  take  the  inductance  in  the  air  space  between 
the  conductors.  Formula  (2)  ;  we  arrive  at  the  values, — 
/-  =  0.0046179  henries  per  mile  of  circuit. 
C  =  0.000  000  006  23  farads  per  mile  of  cir- 
cuit. 


therefore  T''  =  ~  , 

V  ooo,? 6// 9  X  0.000C0000O2J 

186  000  miles  per  second ( 19) 

which  is  approximately  the  spee<.l  of  light. 


CHAPTER  III 
QUICK  ESTIMATING  TABLES 


FOR  every  occasion  where  a  complete  calculation 
of  a  long  distance  transmission  line  is  made,  there 
are  many  where  the  size  of  wire  needed  to  trans- 
mit a  given  amount  of  power  economically  is  required 
quickly.  This  knowledge  is,  moreover,  the  basis  for 
all  transmission  line  calculations,  as  all  methods  of  cal- 
culating regulation  presuppose  that  the  size  of  wire  is 
known.  To  determine  quickly  and  with  the  least  pos- 
sible calculation  the  approximate  size  of  conductor  cor- 
responding to  a  given  I^R  transmission  loss  for  any 
ordinary  voltage  or  distance,  is  the  function  of  Tables 
XII  to  XXI  inclusive.  By  includmg  so  many  trans- 
mission voltages  it  is  not  intended  to  indicate  that  any 
of  them  might  equally  well  be  selected  for  a  new  in- 
stallation. On  the  contrary  it  is  very  desirable  in  the 
consideration  of  a  new  installation,  to  eliminate  con- 
sideration of  some  of  the  voltages  now  in  use.  This 
point  will  be  considered  later. 

Since  both  the  power-factor  of  the  load,  and  the 
charging  current  of  the  circuit,  as  well  as  any  change  in 
the  resistance  of  the  conductors,  will  alter  the  PR  loss, 
it  is  evident  that  it  is  impractical  to  present  tables  which 
will  take  into  account  the  effect  of  all  of  these  variables. 
The  accompanying  tables  do,  however,  give  the  per- 
centage I^R  loss  corresponding  to  the  two  temperatures 
(25  and  65  degrees  C)  ordinarily  encountered  in  prac- 
tice and  the  usual  load  power-factors  of  unity  and  80 
percent  lagging,  upon  which  the  k.v.a.  values  of  the 
tables  are  based.  The  effect,  however,  of  charging  cur- 
rent, corona  or  leakage  loss  is  not  taken  into  account  In 
these  table  values.  The  latter  two  (corona  and  leak- 
age) are  usually  small  and  need  not  be  considered  here. 
The  effect  of  charging  current,  may,  however,  with 
long  circuits  be  material  and  will  be  discussed. 

The  values  of  k.v.a.  in  these  tables  are  based  upon 

the    following    percentage    PR    loss    in    transmission 

(neglecting  the  effect  of  charging  current) : — 

Percent  Percent 

Loss  Loss 

At  25°C  At  6s°C 

Load  at   100  percent  P-F.                 8.66  lo.o 

Load  at     80  percent  P-F.               10.8  12.5 

These  loss  values  are  based  upon  the  power  de- 
livered at  the  end  of  the  circuit  as  100  percent,  and 
not  upon  the  power  at  the  supply  end.  If  raising  or 
lowering  transformers  are  employed,  the  loss  and  vol- 
tage drop  in  them  will,  of  course,  be  in  addition  to  the 
above. 

At  first  glance,  some  of  these  tables  may  appear 
to  have  been  carried  to  extremes  of  k.v.a.  values  for 
the  conductor  sizes.  This  is  because  the  tables  are  cal- 
culated for  ten  percent  loss,   (at   100  percent  power- 


factor  and  65  degrees  C)  whereas  tne  permissible  loss 
is  frequently  much  less  than  ten  percent.  As  the  loss, 
is  directly  proportional  to  the  load,  the  permissible  loads 
for  a  given  size  wire  and  distance  can  be  read  almost 
directly  for  any  loss.  Thus  for  a  two  percent  loss  the 
permissible  k.v.a.  will  be  two-tenths  the  table  values. 
Conversely,  the  size  of  wire  to  carry  a  given  k.v.a. 
load  at  two  percent  loss  will  be  the  same  as  will  carry 
five  (1CK-2)  times  the  k.v.a.  at  ten  percent  loss.  In 
other  words  to  find  the  size  of  wire  to  carry  a  given 
k.v.a.  load  at  any  desired  percent  loss,  find  the  ratio  of 
the  desired  PR  loss  to  the  PR  loss  upon  which  the  table 
values  are  based  (corresponding  of  course  to  the  tem- 
perature and  the  load  power- factor).  Divide  this  ratio 
into  the  k.v.a.  to  be  transmitted.  The  result  will  be  the 
table  k.v.a.  value  corresponding  to  the  desired  PR  loss. 

For  example : — Assume  400  k.v.a.  is  to  be  delivered 
a  distance  of  14  miles  at  6000  volts,  three-phase,  and 
80  percent  power-factor  lagging,  at  an  assumed  temper- 
ature of  25  degrees  C.  Table  XV  indicates  that  this  con- 
dition will  be  met  with  an  PR  loss  of  10.8  percent  if 
No.  O  copper  or  167800  circ.  mil  aluminum  conductors 
are  used. 

Now  assume  that  the  PR  loss  should  not  exceed 
5.4  percent,  in  place  of  10.8  percent  (upon  which  the 
table  values  are  based) .  5.4  -;-  10.8  =  0.5  and  400  -i-  0.5 
=  800  k.v.a.  as  the  table  value  corresponding  to  an  PR 
loss  of  5.4  percent.  The  conductors  corresponding  to 
800  k.v.a.  table  value  (5.4  percent  PR  loss)  will  be 
seen  to  be  No.  0000  copper  or  336420  circ.  mil  alum- 
inum. 

If  conductors  corresponding  to  15  percent  PR 
loss  are  desired  the  same  procedure  will  be  followed : — 
i5-f-io.8  =1.39  and  400-^-I.39  =287  k.v.a.  table  value. 
This  table  value  corresponds  to  approximately  No.  i 
copper  or  133  220  circ.  mil  aluminum  conductors. 

The  table  k.v.a.  values  have  been  tabulated  for  var- 
ious distances.  Should  the  actual  distance  be  different 
from  the  table  values  and  it  is  desired  to  obtain  k.v.a. 
values  corresponding  to  the  losses  upon  which  the  table 
k.v.a.  values  have  been  calculated,  the  following  pro- 
cedure may  be  followed  :— 

For  a  given  PR  loss  in  a  given  conductor  (effect 
of  charging  current  neglected)  the  k.v.a.  X  feet  or  the 
k.v.a.  X  miles  is  a  constant.  Thus  Table  XII  indicates 
that  for  2  000  000  circ.  mil  cable,  756  coo  k.v.a.  X  f^^t 
is  the  constant;  that  is  756  k.v.a.  may  be  transmitted 
1000  feet ;  378  k.v.a.,  2000  feet,  and  so  on.  If  the  actu;'.l 
distance  to  be  transmitted  is  1300  feet  the  correspond- 
ing k.v.a.  value  will  be  756000-7-1300  or  581  k.v.a. 
Usually  the  k.v.a.  value  can  readily  be  approximated 


24 


QUICK  ESTIMATING  TABLES 


for  any  distance  with  sufficient  accuracy  for  the  pur- 
pose for  which  these  quick  estimating  tables  are  pre- 


sented.    One  way  of  dong  this  would  be  as  follows: — 
The  k.v.a.  value  corresponding  to  2500  ft.  is  302  k.v.a. 


TABLE  XII-QUICK  ESTIMATING  TABLE 


CONDUCTORS 

KILOVOLT- AMPERES.  3  PHASE.  WHICH  MAY  BE  DELIVERED  AT  THE  FOLLOWING  VOLTAGES  OVER  THE  VARIOUS 
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The  heating  limitations  may,  for  the  shorter  distances,  particularly  if  insulated  or  concealed  conductors  are  employed,  necessitate  the 
nae  of  larger  conductors,  resulting  in  a  correspondingly  less  transmission  loss.  In  the  case  of  insulated  or  concealed  conductors,  should  the 
It.T.a.  values  fall  near  or  to  the  left  of  the  heavy  line,  consult  Table  XXV"  for  insulated  or  Table  XXIII  for  bare  conductors.  The  reactance  for 
the  larger  tonductors  may  be  excessive,  particularly  for  eOcycIe  service,  producing  excessive  voltage  drop.  This  may  be  obviated  by  installing 
two   or  more  parallel   circuits  or  using  three-conductor   cables.     For   single-phase  circuits  the  k.v.a.  will  be  one-half  the  table  values. 


QUICK  ESTIMATING  TABLES 


^5 


Hence  the  value  corresponding  to  half  this  distance 
(1250  ft.)  is  604  k.v.a.,  which  is  sufficiently  accurate 
for  practical  purposes. 


REACTANCE  LIMITATIONS 

The  k.v.a.  value  of  the  tables  naturally  do  not  take 
into  account  the  reactance  of  the  circuit.     It  will  be 


TABLE  XIII-QUICK  ESTIMATING   TABLE 


CONDUCTORS 

KILOVOLT- AMPERES.  3  PHASE  WHICH  MAY  BE  DELIVERED  AT  THE  FOLLOWING  VOLTAGES  OVER  THE  VARIOUS 
CONDUCTORS  FOR  THE  DISTANCES  STATED.  BASED  UPON  THE  FOLLOWINQ  |2r  LOSS  (EFFECT  OF  CHARGING  CURRENT 
NEGLECTED)                                                                                              AT25"C 

FOR  LOAD  POWEB-FACTOR  OF  IOO%-8  66%  LOSS-  10.0%  LOSS 

i 

05 

m 

COPPER 

AREA 

IN 

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MII.S 

ALUMINUM 

AREA 

IN 

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MILS 

FOR  LOAD  POWER-FACTOR  OF  80%-l0.8%  LOSS-  12  5     LOSr 

550   VOLTS    DELIVERED 

50 

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100 
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The  heating  limitations  may.  tor  the  shorter  distances,  particularly  if  insulated  or  concealed  conductors  are  employed,  necessitate  the 
use  of  larger  conductors,  resulting  in  a  correspondingly  less  transmission  loss.  In  the  case  of  insulated  or  concealed  conductors,  should  the 
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the  larger  conductors  may  be  excessive,  particularly  for  60-cycle  service,  producing  excessive  voltage  drop.  This  may  be  obviated  by  inatalling 
two  or  more  parallel  circuits  or  using  three-conductor  cables.     For  single-phase  circuit*  the  k.v.a.  will  be  one-half  the  table  valuai. 


26 


QUICK  ESTIMATING  TABLES 


necessary  in  some  cases  of  low  voltage  and  single  con- 
ductors^ (where  the  reactance  is  high)  to  use  lower 
values  of  k.v.a.  or  even  in  some  cases  to  multiple  cir- 


cuits in  order  to  keep  the  reactance  within  satisfactory 
operating  limits.  This  will  be  considered  later  by  ex- 
amples on  voltage  regulation. 


TABLE  XIV-QUICK  ESTIMATING  TABLE 


CONDUCTORS 

KILOVOLT-AMPERES.  3  PHASE.  WHICH  MAY  BE  DELIVERED  AT  THE  FOLLOWING  VOLTAGES  OVER  THE  VARIOUS 
CONDUCTORS  FOR  THE  DISTANCES  STATED.  BASED  UPON  THE  FOLLOWING  |2r  LOSS  (EFFECT  OF  CHARGING  CURRENT 
NEGLECTED)                                                                                                       AT26"C            AT  65- O 
FOR  LOAD  POWER-FACTOR  OF  IOO%-8  66%  LOSS-  10.0%  LOSS 

i 

CO 

«1 

CO 

COPPER 

AAEA 

IN- 

OIROUUUt  ' 

MILS 

ALUMINUM 

AREA 

IN 

CIRCULAR 

MIL* 

FOR  LOAD  POWER-FACTOR  OF  80%- 10.8%  LOSS-  12.5     LOSS 

2200  VOLTS    DELIVERED 

100 

FEET 

200 
FEET 

300 
FEET 

500 
FEET 

750 
FEET 

1000 
FEET 

2500 
FEET 

4000 
FEET 

MILE 

MILES 

MILES 

2 

MILES 

22 
MILES 

3 

MILES 

3i 

MILES 

4 

MILES 

5 

MILES 

2  OOO  OOO 
I900  ooo 

/700  ooo 

7S6  000 

6S6000 
641 00c 

a79ooa 

34'3  000 
32000c 

3S1000 

329000 

ZI4000 

ISIOOO 

137000 
129000 

101000 
91  soo 
SSSoo 

7S600 
69600 
6>*  100 

30200 
27400 
2S6ao 

;»Too 
17  100 
16000 

/4300 
/3000 
/z  100 

//  460 
10400 
97IO 

9  sso 

S670 
8100 

7/*0 

«^oo 

*070 

J-730 
S20O 

4  ejo 

4  770 
4330 
4ojro 

4  o?o 
37/0 
3470 

3SSC 

3  2SC 
3  04c 

2  8*0 
2*00 

2430 

/  600  ooo 
/^oo  ooo 
/■foo  ooo 

iOSOOO 

Sttooo 
SZ^ooa 

302000 
293000 

202000 
199000 

176000 

121000 
113000 

106000 

80  700 
7S.sao 
70  .soo 

6  a  SOO 
S6  600 
S2  900 

Z4  200 

22600 
21 10a 

IS  100 
I4IOO 

I3300 

II 400 
10700 
10  000 

9  170 
8S80 
8  010 

7640 
7 /SO 
6,680 

S730 
S3k0 
SOlO 

4S8a 
4290 
4010 

3  82  0 

3.5^70 

J340 

3  2  70 
3  0*0 
2  8*0 

Z  860 
2&SC 
2  SOO 

2290 
Z140 

3  oao 

/zoo  ooo 
1  /oo  ooo 
/  ooo  coo 

/  S9Q  ooo 

4S200O 

417000 
37900a 

226000 
iOfOOC^ 

19^000 

ISIOOO 
131000 
126000 

^0  SOO 
93400 

7S6ao 

60  300 
JS600 
S0400 

4S2O0 
4170a 
378O0 

I8100 

/6700 
/J/OO 

II  300 

/0400 

9  4S0 

8S60 
7900 
7160 

6  SSO 
6320 
S730 

S7/0 
SZ70 
4  770 

42B0 
3  9S0 
3SZ0 

3430 

3  160 
2860 

2  sso 
2630 
2  390 

2  4  SO 
2  2^0 

2  040 

2I40 
1970 
1  790 

mo 
/SSO 
/  430 

9SO  ooo 
900  ooo 
SSO  ooo 

/  S/SOOO 

/43I  ooo 
/3SI  soo 

361 000 
341000 
3  2000c 

ISO  000 
170000 

l&OOOO 

120000 

114  000 

107000 

72200 

i>9  100 

<-»  too 

4SI00 
4S4oa 

42700 

36ioa 

34100 
32  000 

14400 
/3  600 
/2SOO 

9030 

8S30 
SOlO 

6  840 
&4S0 
6070 

S470 
SI60 
4«*0 

4  £60 
4  300 
4  O.S-0 

3420 
3230 
3030 

274° 
ZSSO 
2430 

Z  280 
2  /SO 
2020 

1  9SO 
/  840 
1  73  0 

17 10 
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/  370 
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7SO  ooo 
700  ooo 

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/  I9Z  soo 
/  113  ooo 

joa  000 
ZS^^ooo 
264000 

ISIOOO 

142000 
132000 

101 000 
94  600 
99  100 

^osoo 
S6SOO 
S290O 

40300 
37  SOO 
3S200 

30200 
29400 
2t400 

/2100 
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7S60 
7  100 
6  610 

S730 
S370 
S  010 

4SBO 
4300 
4010 

3»20 
3  SSO 
3340 

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2  TOO 
ZSOO 

3290 
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1910 
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/  670 

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1430 
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1  2Sa 

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1  070 
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/033  soo 

9S+  ooo 

S74  soo 

2A60O0 
2  26000 

20900c 

12300C 
113  000 

10400a 

91900 
7S400 
69  SOO 

49100 

4i200 
4/700 

33700 
30  100 
2  7»O0 

24  SOO 
Z26aa 
20  Sao 

9  800 

90SO 

8  3SO 

6  140 
S6SO 
S  210 

4  6S0 
4  ZSO 

3  9SO 

3720 
343a 
3  I60 

3100 
2  SSO 
2  630 

2  3  30 
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/  y7o 

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1  7/0 
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/330 
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1070 

987 

931 

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790 

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4-00  ooo 

79S  ooo 
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IS1 000 

170000 
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99  soo 

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7SSO0 

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S6  600 

So  600 

3  7  SOO 
33900 
30  30c 

2S200 
22*00 
20200 

1S900 
17000 
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7S60 
6,800 
6  070 

4730 
4  240 
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3  SSO 

3220 
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Z  140 
1  910 

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1  610 
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1  190 

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824 

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4.77000 

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132000 

113000 

94100 

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44  100 
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31  600 

24+00 
22700 
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/s  100 

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n  300 

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3  800 

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336  4ZO 
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80000 

63700 
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40000 
31  900 
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2*700 
21  200 
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12.700 
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10700 
9490 
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8  oao 
6  370 
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/  S90 
/  260 

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760 

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636 

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477 

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483 
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403 

318 

433         378 
3+i         301 
273         238 

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8000 
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1  000 

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7,90 
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Vol 
47i 

606 
480 
3  80 

400 
317 

3  80 
300 
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303 
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190 

2S3 
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189 

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41  73» 
33  OSS 

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66  370 
SZ  6  30 

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9900 
7910 
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6  610 

s  270 

4  190 

3970 

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/  670 

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790 
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313 

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4000  VOLTS 

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1 

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li 

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2 

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22 
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3 
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3i 
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4 

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6 

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8 
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A  so  OOO 

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/  033 000 

9S4  000 
S74  soo 

1S400 
lA  200 
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10  2SO 
9460 
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7 100 

6,  sso 

6  ISO 
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4370 

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3S.J-0 
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3280 

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ZS40 
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ZS60 
2370 
21  so 

3  200 
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/920 

/77S 
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/S40 
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l-^oa 
/290 
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II 00 
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79s  000 

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9490 

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7/20 
6330 

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4  7sa 

4  7-ro 
4170 
3  800 

3  960 
3S60 
316S 

3400 
3060 
2713 

Z980 
Z670 
Z373 

Z390 
2/40 
1900 

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1  3S7 

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1666 
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313 

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1970 
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1  670 
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1000 
790 
627 

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62S 
493 
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416 
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3/2 

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277 
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179 
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393 

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199 
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98 

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99 
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4-400   VOLTS    DELIVERED 

1 

MILE 

MILES 

2 

MILES 

2i 
MILES 

3 

MILES 

3i 

MILES 

4 

MILES 

5 
MILES 

6 
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7 
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8 
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9 

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10 

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12 
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13 
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14 
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/  033  000 

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9300 
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4  880 
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4270 
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3720 
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31  00 
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2660 
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2270 

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1990 

2070 
1900 

1770 

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79S  000 
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636 000 

14300 
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633 

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214 
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51 

11 

The  heating  limitations  may,  for  the  shorter  distances,  particularly  if  insulated  or  concealed  eonductore  »re  employed,  necasstttte  the 
see  of  larger  conductors,  resulting  in  a  correspondingly  less  transmission  loss.  In  the  case  of  insulated  or  concealed  conductors,  should  the 
k.T.a.  Talues  fall  near  or  to  the  left  of  the  heavy  line,  consult  Table  XXV  for  insulated  or  Table  XXIII  for  bare  conductors.  The  reactance  for 
the  larger  conductors  may  be  excessive,  particularly  for  60-cycle  service,  producing  excessive  voltage  drop.  '  This  may  be  obviated  by  installin( 
two  or  more  parallel   circuits  or  using  three-conductor   cables.     For  ■iscle-phase  circuits  the  k.y.k.  will  be  one-halt  the  table  value*. 


QUICK  ESTIMATING  TABLES 

TABLE  XV-QUICK  ESTIMATING  TABLE 


27 


CONDUCTORS 

KILOVOLT- AMPERES.  3  PHASE.  WHICH  MAY  BE  DELIVERED  AT  THE  F 
CONDUCTORS  FOR  THE  DISTANCES  STATED.  BASED  UPON  THE  FOLLOWING 
NEGLECTED) 

FOR  LOAD  POWER-FACTOR  OF  I00*-856*r7§a_ 

OLLOW 

i'rlc 

AT  66 
10.0%  L 
12.6     L 

D 

NO  VOLTAGES  OVER  THE  VARIOUS 
ISS  (EFFECT  OF  CHARGING  CURRENT 

•c 

d 

z 

CO 

COPPER 

AREA 

IN 

CIRCULAR 

MIL« 

ALUMINUM 

AREA 

tN 

CIRCULAR 

MIL* 

FOR  LOAD  POWER-^^ACTOR  OF  80%- 10.8%  LOSS- 

OSS 
3SS 

6000   VOLTS    DELIVERE 

1 
MILE 

MILES 

2 

MILES 

2x 

MILES 

3 

MILES 

3i 

MILES 

4 

MILES 

5 

MILES 

6 

MILES 

7 
MILES 

8 

MILES 

9 

MILES 

10           II            12           13           14 
MILES  MILES   MILES  MILES  MILES 

eoQooa 
j-soooo 

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3 1  eoo 

2  9S0C 

33100 
31  300 

19  700 

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4220 

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3980    3S90    3180 
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3 'SO       2890    2470 
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12470 
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The  heating  limitations  may.  for  the  shorter  distances,  particularly  if  insulated  or  concealed  conductors  are  employed,  necessitate  tha 
ase  of  larger  conductors  resulting  id  a  correspondingly  less  transmission  lo8$  In  the  case  of  insulated  or  concealed  conductors  should  the 
k.v.a  valueir  fal'  near  or  to  the  left  of  the  heavy  line  consult  Table  XXV  for  insulated  or  Table  XXIII  for  bare  conductors  The  reactance  lor 
the  larger  conductors  may  be  excessive  particularly  for  60-cycle  service,  producing  excessive  voltage  drop  This  may  be  obviated  by  installinf 
two  or   more   parallel   circuits   or   using   three-conductor   cables.      For   single-phase  circuits  the  k.v.a.  will   be  one-half  the  table  values 


QUICK  ESTIMATING  TABLES 


TABLE  XVI-QUICK  ESTIMATING  TABLE 


CONDUCTORS 

KILOVOLT- AMPERES.  3  PHASE.  WHICH  MAY  BE  DELIVERED  AT  THE  FOLLOWING  VOLTAGES  OVER  THE  VARIOUS 
CONDUCTORS  FOR  THE  DISTANCES  STATED.  BASED  UPON  THE  FOLLOWING    i2r  LOSS  (EFFECT  OF  CHARGING  CURRENT 

FOR  LOAD  POWER-FACTOR  OF  I00%-8.66%  LOSS-  10.0%  LOSS 

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,^  5    ''""*a''oi8    may,    for    the    shorter    distances,    particularly  if   insulated   or  concealed   conductors  are  employed,   necessitate  the 

ase  01  larirer  cou/luctors.  resultin?  in  a  correspondingly  less  transmission  loss.  Tn  the  case  of  insulated  or  concealed  conductors  should  ihfl 
Kv.a  values  fall  near  or  to  the  left  of  the  heavy  line,  consult  Table  XXV  for  insulated  or  Table  XXIII  for  bare  conductors.  The  reactance  for 
ine  larcer  conductors  may  he  excessive,  particularly  for  60-cycle  service,  producing  excessive  voltage  drop.  This  may  be  obviated  by  installing 
two   or  more   parallel    circuits   or   usine   ihreecondurtor   cables.      For   sinrle-phRse  circuits  the  k.v.a.   will  be  one-half  the  table  values. 


QUICK  ESTIMATING  TABLES 


i9 


TABLE  XVII-QUICK  ESTIMATING  TABLE 


CONDUCTORS 

KILOVOLT- AMPERES.  3  PHASE.  WHICH  MAY  BE  DELIVERED  AT  THE  FOLLOWING  VOLTAGES  OVER  THE  VARIOUS 
CONDUCTORS  FOR  THE  DISTANCES  STATED.  BASED  UPON  THE  FOLLOWING  |2rlOSS  (EFFECT  OF  CHARGING  CURRENT 
NEGLECTED) 

FOR  LOAD  POWER-FACTOR  OF  IOO*-8.66%  LOSS-  lOOTb^^ 
FOR  LOAD  POWER-FACTOR  OF  80«^I0,8%  LOSS-  12  5     LOSS 

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The  heating  limitations  may,  for  the  shorter  distances,  particularly  if  insulated  or  concealed  conductors  are  employed,  necessitate  th» 
ase  of  larger  conductors,  resulting  in  a  correspondingly  less  transmission  loss.  In  the  case  of  insulated  or  concealed  conductors,  should  the 
k.v.B.  values  fall  near  or  to  the  left  of  the  heavy  line,  consult  Table  XXV  for  insulated  or  Table  XXIII  for  bare  conductors.  The  reactance  for 
the  larger  conductors  may  bo  excessive,  particularly  for  eo-cycle  service,  producing  excessive  voltage  drop.  This  may  be  obviated  by  instaUiii( 
two  or  more  parallel  circuits  or  using  three-conductor  cables.     For  Bin;le-phase  circuits  the  k.v.a.  will  be  one-half  the  table  values. 


30 


QUICK  ESTIMATI.W.  TABLES 


TABLE  XVIM-QUICK  ESTIMATING  TABLE 


CONDUCTORS 

KILOVOLT- AMPERES.  3  PHASE.  WHICH  MAY  BE  DELIVERED  AT  THE  FOLLOWING  VOLTAGES  OVER  THE  VARIOUS 
CONDUCTORS  FOR  THE  DISTANCES  STATED.  BASED  UPON  THE  FOLLOWING   |2r  LOSS  (EFFECT  OF  CHARGING  CURRENT 
NEGLECTED)                                                                                                 Ajjg.c           AT66°0 

FOR  LOAD  POWER-FACTOR  OF  IOO%-8.66%  LOSS-  10.0%  LOSS 

FOR  LOAD  POWER-FACTOR  OF   80%- 10.8%  LOSS-  12.6     LOSS 

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The  loss  due  to  corona  will  not  be  excessive  with  any  of  the  above  conductors  used  at  sea  level  for  the  voltages  stated.  For  elevations 
above  sea  level,  check  the  values  with  Table  XXII,  especially  for  the  smaller  conductors.  On  long  circuits  of  high  voltage,  the  effect  of 
charging  current  (also  corona  and  leakage  losses)  will  be  to  increase  or  decrease  the  I'R  loss,  depending  on  the  amount  of  load  and  its  power- 
factor.     See  Pig.   18 


QUICK  ESTIMATING  TABLES 

TABLE  XIX-QUICK  ESTIMATING  TABLE 


31 


CONDUCTORS 

KILOVOLT- AMPERES.  3  PHASE,  WHICH  MAY  BE  DELIVERED  AT  THE  FOLLOWING  VOLTAGES  OVER  THE  VARIOUS 
CONDUCTORS  FOR  THE  DISTANCES  STATED.  BASED  UPON  THE  FOLLOWING    |2r  LOSS  (EFFECT  OF  CHAROINO  CURRENT 
NEGLECTED)                                                                                                      AT26'C 

FOR  LOAD  POWER-FACTOR  OF  I00%^8.66%  LOSS-  IO.O%^LOSS 
FOR  LOAD  POWER-FACTOR  OF  80%^I0.8%  LOSS-  12  6     LOSS 

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The  loss  due  to  corona  will  not  be  excessive  with  any  of  the  above  conductors  used  at  nea  level  for  the  voltaRes  slated.     For  elevatioDi 

above   ses  level,   checlc  the   values   with  Table   XXII.   especially  for  the  smaller     conductors.       On    lonK    circuits    of    high    voltaje,    the    eflFecl    of 

charging  current    (also  corona  and  leakage  losses)    will  be  to  increase  or  decrease  the  I  R  loss,  depending  on  the  amount  of  load  and  its  power- 
factor.     See  Fig.   13 


32 


QUICK  ESTIMATING  TABLES 

TABLE  XX-QUICK  ESTIMATING  TABLE 


CONDUCTORS 

KILOVOLT- AMPERES.  3  PHASE.  WHICH  MAY  BE  DELIVERED  AT  THE  FOLLOWING  VOLTAGES  OVER  THE  VARIOUS 
CONDUCTORS  FOR  THE  DISTANCES  STATED.  BASED  UPON  THE  FOLLOWINQ  (Sr  lqsS  (EFFECT  OF  CHARGING  CURRENT 

f^E^^EC™                                                                                                       AT25»C            ATefi«C 
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The  loss  due  to  corona  will  not  be  excessive  with  any  of  the  above  conductors  used  at  sea  level  for  the  voltages  stated.  For  elevations 
above  sea  level,  check  the  values  with  Table  XXII,  especially  for  the  smaller  conductors.  On  long  circuits  of  high  voltage,  the  effect  of 
charging  current  (also  corona  and  leakage  losses)  will  be  to  increase  or  decrease  the  I'B  loss,  depending  on  the  amount  of  load  and  its  power- 
factor.     See  Fin.   13 


QUICK  ESTIMATING  TABLES 


33 


HEATING  LIMITATIONS 

The  k.v.a.  values  given  in  these  tables  do  not  take 
into  account  the  heating  and  consequently  carrying  cap- 
acity of  the  conductors.  This  may  be  ignored  in  the 
case  of  the  longer  overhead  high-voltage  transmission 
circuits.  For  very  short  circuits  (especially  for  the 
lower  voltages  and  particularly  for  insulated  or  con- 
cealed conductors)  the  carying  capacity  (safe  heating 
limits)  of  the  conductors  must  be  carefully  considered. 


approximately  the  point  at  which  the  carrying  capacity 
of  that  particular  conductor  is  reached  if  insulated  and 
installed  in  a  fully  loaded  four  duct  line.  If  the  con- 
ductor is  to  be  installed  in  a  duct  line  having  more  than 
four  ducts  its  Opacity  will  be  still  further  reduced. 
The  position  of  this  line  is  based  upon  the  use  of  lead 
covered,  paper  insulated,  three  conductor,  copper  cables 
for  sizes  up  to  700000  circ.  mils  and  of  lead  covered, 
paper  insulated,  single  conductor,  copper  cables  for  the 
larger  sizes.     In  other  words,  the  position  of  this  heavy 


TABLE  XXI-QUICK  ESTIMATING  TABLE 


CONDUCTORS 

KILOVOLT.  AMPERES.  3  PHASE  WHICH  MAY  BE  DELIVERED  AT  THE  FOLLOWING  VOLTAGES  OVER  THE  VARIOUS 
CONDUCTORS  FOR  THE  DISTANCES  STATED.  BASED  UPON  THE  FOLLOWING   |2r  LOSS  (EFFECT  OF  CHARGING  CURRENT 

FOR  LOAD  POWER-FACTOR  OF  IOO%-8  66%  LOSS-  10  0%  LOSS 
FOR  LOAD  POWER-FACTOR  OF   8CWfr- 10.8%  LOSS- 12.5     LOSS 

COPPER 

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187,000   VOLTS   DELIVERED 

06 
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104 

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112 

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120 
MILES 

128 
MILES 

144 

MILES 

160 

MILES 

176 

MILES 

192 
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112 
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128 
MILES 

144 

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160 
MILES 

176 

MILES 

192 
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208 

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224 
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240 

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The  loss  due  to  corona  will  not  be  excessive  with  any  of  the  above  condnctors  used  at  sea  level  for  the  voltages  stated.     For  elevatioui 

above   sea  level,    check  the  values  witli   Table  XXII,   especially  for  the  smaller     conductors.       On    long    circuits    of    hijrh    yoitatre,    the    effect    of 

charging  current    (also  corona  and  leakage  losses)   will  be  to  increase  or  decrease  the  I'R  loss,  depending  on  the  amount  of  load  and  its  power- 
factor.     See  Fig.   13 


For  circuits  of  short  length  the  carrying  capacity  of 
conductors  will  frequently  determine  these  sizes  and 
not  the  economic  transmission  loss.  The  carrying  cap- 
acity of  bare  copper  conductors  suspended  in  air  and 
of  insulated  copper  conductors  in  duct  lines  are  given 
in  tables  XXIII  and  XXIV,  both  of  which  are  to  appear 
in  subsequent  articles. 

Running  diagonally  across  each  table  from  XII  to 
XVII  inclusive,  is  a  heavy  line.  The  point  at  which 
this  heavy  line  intersects  the  horizontal  line  containing 
the  k.v.a.  values  for  a  given  size  of  conductor  indicates 


line  is  based  upon  the  k.v.a.  values  for  carrying  capacity 
given  in  Table  XXIV  and  is  placed  upon  the  tables  as  a 
warning  that  the  heating  limit  capacity  of  the  conductors 
must  be  considered.  To  illustrate,  suppose  220  volts  is 
to  be  delivered,  over  i  000  cxx)  circ.  mil,  insulated,  single 
conductor,  copper  cables  in  a  fully  loaded  four  duct 
conduit.  Table  XII  indicates  that  189  k.v.a.  can  be 
transmitted  over  these  conductors  a  distance  of  2000  ft. 
without  overheating  the  cable.  If  it  is  desired  to  trans- 
mit 378  k.v.a.  a  distance  of  1000  feet,  the  fact  that  this 
value  occurs  to  the  left  of  the  heavy  line,  indicates  that 


34 


QUICK  ESTIMATING  TABLES 


it  is  beyond  the  safe  carrying  capacity  for  this  size  con- 
ductor in  a  four  duct  line.  Reference  to  Table  XXIV 
will  show  that  297  k.v.a.  is  the  maximum  capacity  of 
this  cable  under  the  conditions  stated.  In  this  case, 
either  a  larger  conductor,  or  two  or  more  smaller  con- 
ductors must  be  used  to  prevent  overheating.  This  will 
result  in  a  less  loss  than  those  upon  which  the  table 
k.v.a.  values  are  based,  and  in  this  case  the  heating  of 
the  cable  will  probably  determine  the  size  to  use. 

EFFECT  OF  CHARGING  CURRENT  IN  ABOVE  I^R  LOSS  VALUES 

As  stated  previously,  the  percent  PR  losses  in  the 
quick  estimating  tables  are  based  upon  the  load  current 
and  therefore  do  not  take  into  account  the  effect  of  the 
charging  current  which  is  of  a  distributed  nature  and 
superimposed  upon  the  load  current.  The  effect  of  the 
charging  current  is  to  increase  or  decrease  the  current 
in  the  circuit  by  an  amount  depending  upon  the  relative 


2800 


SUPPLY  END 


200  300 

DISTANCE  IN  MILES  FROM  SUPPLY  END 


FIG.    13 — EFFECT  OF  CHARGING  CURRENT  ON   TR  TRANSMISSION   LOSS 

The  curves  represent  (for  certain  circuits)  an  approximation  of  the  resultant  I'R  loss,  compared  to 
what  it  would  have  been  if  there  were  no  charging  current  present  in  the  circuit.  The  effect  of  the 
charging  current  superimposed  upon  the  receiver  current  is  either  to  increase  or  to  decrease  the  PR 
loss  of  the  circuit  depending  principally  upon  the  relative  amount  of  the  leading  and  lagging  components 
of  the  current  in  the  circuit. 

values  of  the  lagging  and  leading  quadrature  compon- 
ents of  the  current  in  the  circuit. 

For  instance  assume  that  the  power-factor  of  the 
load  is  unity.  In  such  case  there  is  no  quadrature  com- 
ponent in  the  load  current.  If,  however,  the  circuit  is 
of  considerable  length,  and  particularly  if  the  frequency 
is  60  cycles,  there  will  be  an  appreciable  amount  of 
charging  current  (quadrature  leading  component) 
added  vectorially  to  the  load  current.  The  sum  of 
these  two  currents  in  quadrature  with  each  other  wiil 
result  in  an  increase  of  current  in  the  circuit  with  a 
consequent  increase  in  the  I^R  loss.  Thus  the  effect 
of  charging  current  in  a  circuit  delivering  a  load  of  100 
percent  power-factor  will  always  be  to  increase  the  PR 
loss. 

If,  however,  the  power- factor  of  the  load  is  laggin(». 


there  will  be  a  lagging  component  in  the  load  current. 
The  charging  or  leading  current  v,-ill  be  practically  in 
opposition  to  the  lagging  component  of  the  load  current 
and  will  therefore  tend  to  cancel  or  neutralize  the  lag- 
ging component  of  the  load  current.     The  result  will 
be  a  reduction  of  the  current  in  the  circuit  and  con- 
sequently in  the  PR  loss.     But  if  the  circuit  is  very 
long,  particularly  if  the  frequency  is  60  cycles  and  the 
load  power-factor  is  near  unity  (lagging  component  in 
load   current   small)    the   comparatively  large   leading 
component  (charging  current)  will  not  only  neutralize 
'the   lagging  component   of  the   load   current,   but   will 
produce  a  leading  power-factor  at  points  along  the  cir- 
cuit.    If  the  charging  current  is  sufficiently  high  it  will 
ncrease  the  current,  causing  an  increase  in  the  PR 
OSS.     Thus  the  effect  of  charging  current  in  circuits 
lelivering  a  lagging  load  is  to  decrease  the  PR  loss  up 

to  a  certain 
amount  and  then, 
if  the  charging 
current  is  s  u  ffi- 
ciently  large,  to  in- 
crease PR  loss. 

The  curves  in 
Fig.  13  show  this 
effect  for  25  and 
60  cycle  circuits 
delivering  loads  of 
unity  power-fac- 
tor; also  loads 
of  80  percent  lag- 
ging power-factor 
for  circuits  up  to 
500  miles  long.  It 
will  be  seen  that 
for  circuits  300 
miles  long  the 
effect  of  charging 
current  will  be  to 
reduce  the  PR 
loss  by  approxi- 
mately 25  percent 
if  the  load  is  80 
percent  lagging.  If  the  load  power-factor  is  unity  the 
PR  loss  will  be  increased  approximately  10  percent  for 
these  particular  problems  if  the  frequency  is  25,  and 
30  percent  if  the  frequency  is  60  cycles. 

The  curves  in  Fig.  13  show  that  for  circuits  500 
miles  long,  in  which  the  entire  charging  current  is  fur- 
nished from  one  end  of  the  circuit,  the  effect  of  this 
charging  current  is  to  increase  the  PR  loss  by  300  per- 
cent if  the  frequency  is  60  cycle  and  the  load  power- 
factor  100  percent.  In  other  words  a  large  part  of  the 
current  in  the  circuit  for  such  a  long  60  cycle  circuit 
is  charging  current  so  that  the  effect  of  the  load  current 
on  the  PR  loss  is  comparatively  small.  Of  course  such 
a  long  circuit,  unless  fed  from  two  or  more  generating 
stations  located  at  widely  separated  points  along  the 
transmission  lin3,  would  not  be  commercially  practical. 


CHAPTER  IV 
CORONA   EFFECT 


In  1898  Dr.  Chas.  F.  Scoft  presented  a  paper  before  the  A.I.E.E.  describing  experimental  tests  (made 
during  several  years  previous)  relating  to  the  energy  loss  between  conductors  due  to  corona  effect.  These 
investigations  began  at  the  Laboratory  at  Pittsburgh  and  were  continued  at  Telluride,  Colorado,  in  con- 
junction with  the  engineers  of  the  Telluride  Power  Company.  Preliminary  observations  were  made  by 
Mr.  V.  G.  Converse  and  were  continued  in  notable  measurements  by  Mr.  R.  D.  Mershon.  These  investiga- 
tions were  later  followed  by  the  work  of  Professor  Ryan,  by  Mr.  R.  D.  Mershon,  Mr.  F.  W.  Peek,  Jr.,  Dr. 
J.  B.  Whitehead,  Mr.  G.  Faccioli  and  others.  The  electrical  profession  is  particularly  indebted  to  Mr.  Peek 
and  Dr.  Whitehead  for  the  large  amount  of  both  practical  and  theoretical  data  which  they  have  presented 
to  the  electrical  profession  on  the  subject.  Mr.  Peek  developed  and  presented  the  empirical  formulas 
which  follow,  for  determining  the  disruptive  critical  voltage,  the  visual  critical  voltage  and  the  power 
loss  due  to  corona  effect.  The  close  accuracy  of  Mr.  Peek's  formulas  has  been  confirmed  by  various  in- 
vestigators in  different  sections  of  the  country.  The  following  deductions  concerning  corona  have  to  a 
large  extent  been  previously  presented  by  Mr.  Peek. 


CORONA,  manifesting  its  presence  usually  by  an 
electrostatic  glow  or  luminous  discharges,  and 
audibly  by  a  hissing  sound,  was  clearly  observed 
and  studied  in  connection  with  electrostatic  machines. 
It  did  not  become  a  serious  factor  to  be  considered  in 
connection  with  the  design  of  commercial  electrical  ap- 
paratus until  the  increasing  generator  and  transmission 
voltage  emphasized  its  important. 

Although  it  is  usual  to  think  of  corona  effect  only 
in  connection  with  high-voltage  transmission  lines,  it 
has  received  not  a  little  thought  of  late  by  the  designers 
of  high-voltage  generators  and  motors,  notably  large, 
high-voltage  turbogenerators.  By  effectively  insulating 
the  portion  of  the  conductor  embedded  in  the  iron  of 
the  armatures  of  alternating-current  machines,  particu- 
larly with  mica,  punctures  to  ground  due  to  corona  ef- 
fect are  not  likely  to  occur.  However,  at  the  ends  of 
the  armature  coils  (where  it  is  difficult  to  employ  mica 
for  insulating),  where  air  is  partially  depended  upon 
as  an  insulating  medium  between  coils  and  ground, 
corona  may  appear.  The  presence  of  these  corona 
stresses  results  in  disintegrating  and  weakening  some 
kinds  of  insulating  materials,  causing  them  to  break 
down  after  a  period  of  service.  This  deterioration  of 
insulation  may  be  due  to  local  heating,  mechanical  vi-. 
bration  or  chemical  formations  in  the  overstressed  air, 
such  as  ozone,  nitric  acid,  etc. 

Higher  voltages  are  being  chosen  as  an  economic 
means  for  reducing  loss  in  transmission.  These  higher 
voltages  may  result  in  corona  loss  far  in  excess  of  the 
saving  in  transmission  loss  due  to  the  adaptation  of  the 
higher  voltages.  It  is,  therefore,  pertinent  that  par- 
ticular consideration  be  given  to  the  limitation  of  corona 
loss  when  the  choice  of  conductors  is  made.  This  con- 
sideration will  sometimes  make  it  desirable  to  take  ad- 
vantage of  the  higher  critical  voltage  limits  of  aluminum 
conductors  (with  steel  reinforced  centers)  of  an  equiva- 
lent resistance,  due  to  their  greater  diameter;  or  it  may 
be  desirable  to  obtain  the  necessary  larger  diameter  by 
the  use  of  copper  conductors  having  some  form  of  non- 
conducting centers   or,   for  still   larger  diameters,   of 


aluminum  conductors  having  such  centers,  in  order  to 
avoid  skin  effect.  The  use  of  copper  conductors  having 
hemp  centers  has  in  some  instances  given  mechanical 
trouble. 

The  critical  voltage  at  which  corona  becomes  mani- 
fest, is  not  constant  for  a  given  line,  but  is  somewhat 
dependent  upon  atmospheric  conditions.  Assuming  a 
Ime  employing  conductors  just  within  the  critical  volt- 
age limitations  for  the  conditions  to  be  met,  the  corona 
loss  in  such  a  line  would  be  almost  negligible  during  fair 
weather,  but  during  stormy  weather,  (particularly  dur- 
ing snowstorms)  this  corona  loss  would  be  many  times 
what  it  is  during  fair  weather.  On  the  other  hand, 
since  the  storm  will  usually  not  appear  over  the  whole 
length  of  lines  at  the  same  time  and  since  storms  occur 
only  at  intervals,  it  may  often  be  economical  to  allow 
this  loss  to  reach  fairly  high  values  during  storms.  Fog, 
sleet,  rain  and  snowstorms  lower  the  critical  voltage 
and  increase  the  losses.  The  effect  of  snow  is  greater 
than  any  other  weather  condition.  Increase  in  tem- 
perature or  decrease  in  barometric  pressure  lowers  the 
voltage  at  which  visual  corona  starts. 

The  critical  voltage  increases  with  both  the  dia- 
meter of  conductors  and  their  distance  apart.  This 
sometimes  makes  it  desirable  to  use  aluminum  conduc- 
tors as  previously  stated.  It  also  increases  with  the 
horizontal  or  vertical  arrangement  of  conductors,  due 
to  the  fact  that  the  two  outside  conductors  considered 
as  a  pair  are  twice  as  far  apart  as  are  the  other  pairs. 
The  same  general  rules  apply  to  stranded  conductors  as 
to  solid  conductors,  the  actual  diameter  of  the  former 
being  considered  as  the  effective  diameter  of  the  con- 
ductor. 

The  losses  due  to  corona  effect  increase  very 
rapidly  with  increa.se  in  voltage  after  the  critical  voltage 
has  been  reached.  A  long  transmission  line  having 
considerable  capacitance  may  deliver  a  higher  voltage 
than  appears  at  the  generator  end  of  the  fine  due  to 
capacitance  effect.  The  corona  loss  would  in  this  case 
be  greater  per  mile  at  the  receiving  end  than  at  the 
sending  end  of  the  line. 


36 


CORONA  EFFECT 


The  magnitude  of  the  losses,  as  well  as  the  critical 
voltage,  is  affected  by  atmospheric  conditions; — hence 
tliey  probably  vary  with  the  particular  locality  and  the 
season  of  the  year.  Therefore,  for  a  given  locality,  a 
voltage  which  is  normally  below  the  critical  point,  may 
at  times  be  above  the  critical  voltage,  depending  upon 
changes  in  the  weather. 

The  material  of  the  conductors  does  not  seem  to 
affect  the  losses.  Sometimes  the  conductors  of.  new 
transmission  lines,  when  first  placed  in  service  will  show 
visual  corona,  which  may  entirely  disappear  after  a  few 
hours  or  weeks  of  service.  This  may  be  due  to 
scratches,  particles  of  foreign  substances,  etc.,  on  the 
conductors  which  are  eliminated  after  the  voltage  stress 
has  been  kept  on  the  conductors  for  a  short  time.  Un- 
der such  conditions  the  corona  loss  will  also  become 
less  as  the  visual  effect  disappears. 

The  loss  of  power  due  to  corona  effect  increases 
with  frequency  and  increases  as  the  square  of  the  ex- 
cess voltage  above  a  certain  critical  voltage  referred  to 
as  the  "disruptive  critical  voltage"  e^.  This  disruptive 
critical  voltage  is  that  voltage,  at  which  a  certain  definite 
and  constant  potential  gradient  is  reached  at  the  con- 
ductor surface.  This  gradient  go  is  30  kv  maximum 
(21. 1  kv  effective)  per  centimeter,  or  76.2  kv  maximum 
(53.6  kv  effective)  per  inch.  These  values  are  based 
upon  an  air  density  at  sea  level  (25°  C,  29.92  inches  or 
76  cm.  barometer).  This  gradient  is  independent  of 
the  size  of  conductors  and  thcr  distance  apart,  but  is 
proportional  to  the  air  density,  that  is  to  the  barometric 
pressure  and  the  absolute  temperatures.  It  may  be  con- 
sidered as  the  dielectric  strength  of  air.  The  presence 
of  corona  at  a  certain  point  of  the  system  shows  that  a 
critical  electric  stress  has  been  exceeded  at  that  point. 
The  corona  loss  is  also  proportional  to  the  square  root 
of  the  conductor  radius  r  and  inversely  proportional  to 
the  square  root  of  the  conductor  spacing. 

The  law  by  which  corona  losses  increase  with  the 
voltage  does  not  give  a  very  steep  curve,  but  a  rather 
mild  curve  following  the  quadratic  law  at  and  above  the 
critical  limit.  In  other  words  there  is  no  sharp  elbow 
in  the  curve  above  which  the  losses  increase  very  rapidly 
with  the  voltage  and  which  could  be  adopted  as  the 
normal  operating  point  of  the  circuit. 

Table  XXII,  indicating  the  voltage  limitations  due 
to  corona  effect,  has  been  worked  up  from  Mr.  F.  W. 
Peek's  formula  as  indicated  at  the  bottom  of  the  table. 
The  values  in  this  table  are  conservative  and  may  in 
many  cases  be  exceeded.  They  are  the  effective  e^  dis- 
ruptive critical  voltage  between  conductors  for  fair 
weather  based  upon  S  values  for  25  degrees  C.  (^y 
degrees  F)  and  m^  values  of  0.87  for  cable  and  0.93 
for  wire.  With  these  table  values,  corona  loss  should 
not  be  excessive  during  storms.  If  the  values  of 
Table  XXII  indicate  that  the  conductors  contemplated 
are  close  to  the  limit  due  to  corona  effect,  a  careful 
check  should  be  made  by  the  formula  to  determine 
definitely  the  corona  loss  for  such  conductors  under 
storm  operating  conditions. 


e,  =  2.302  7>H  go  hr 


(-^) 


logv 


.  {21) 


(e  —  Co)  ^  lo--' {22) 


F.   W.   PEEK  S  CORONA  FORMULAE 

Disruptive  Critical  Volts,  Fair  Weather  (parallel  wires) 

fo  =  2.302  niogo  8  r  logm—  {20) 

effective  kv  to  neutral, — 
Visual  Critical  Volts — Fair  Weather   (parallel  wires) 

s 
r 

effective  kv  to  neutral 
Power  Loss   (fair  weather)  — 

P -'f  {/ +  ^5)  ^T 

kw  per  mile  of  each  conductor 

Power  Loss  {Storm) — Storm  power  loss  is  higher  and  can 
generally  be  found  with  fair  approximation  by  assuming  <?<,  ^= 
0.80  times  fair  weather  c„.  It  generally  works  out  in  practice 
that  the  Co  voltage  is  the  highest  that  should  be  used  on  trans- 
mission  lines    {22A ) 

All  of  the  above  voltages  are  to  neutral.  To  find  voltages 
between  lines  multiply  by  1.73  for  three-phase,  and  by  2  for 
single  phase. 

Notation — 

e     =  Effective  applied  voltage  in  kv  to  neutral. 

(This  will  vary  at  different  points  of  the  circuit  and 
at  different  loads.  At  low  loads  and  long  lines  of 
high  voltage  it  may  be  higher  at  the  receiving  end 
than  at  the  generator  end  due  to  inductive  capaci- 
tance) 

fo  =  effective  disruptive  critical  voltage  in  kv  to  neutral. 
It  is  the  voltage  that  gives  a  constant  break  down 
gradient  for  air  of  76  kv  maximum  per  inch,  the 
"elastic  limit"  at  which  the  air  breaks  down. 
Visual  corona  does  not  start  at  the  disruptive 
critical  voltage,  but  at  a  higher  voltage  Cv. 

Cf  =  effective  visual  critical  kv  to  neutral  (voltage  at 
which  visual  corona  starts) 

P  =  power  loss  in  fair  weather  in  kw  per  mile  of  single 
conductor, 

J  7.9b 


s     = 


This  takes  care  of  the  effect  of  altitude 


459  +  t 

and  temperature,   (air  density).     It  is  I  at  25  de- 
grees C.  {77  degrees  F.)  and  29.92  inches  (76  cm.), 
barometric  pressure. 
(7o   =  53.6  kv  per  inch  effective  (disruptive  gradient  of  air) 
b     =:  barometric  pressure  in  inches. 
t      =  maximum  temperature  in  degrees  F. 
/     =  frequency  in  cycles  per  second, 
m.  =  irregularity  factor. 
=  I  for  polished  wires. 

=^  0.98  to  0.93  for  roughened  or  weathered  wire. 
=  0.87  to  0.83  for  cables. 
»ir  =  nto  for  wires  (l  to  O.93) 

=  0.72  for  local  corona  all  along  cables   (7  strands) 
:=  0.82  for  decided  corona  all  along  cables  (7  strands) 
r     r=z  radius  of  conductor  in  inches. 

J     =  spacing  in  inches  between  conductor  centers,  based 
upon  the  assumption  of  a  symmetrical  triangular 
arrangement.     For  three-phase  irregular  flat  or  tri- 
angular spacing  take  .r  :=  f  Anc        .    For  three- 
phase  regular  flat  spacing  tako  s  =  J.26A. 
Theoretically,  if  the  conductors  were  perfectly  smooth,  no 
loss  would  occur  until  the  critical  voltage,  Ct  is  reached,  when 
the  loss  should  suddenly  take  a  definite  value,  equal   to  that 
calculated  by  quadratic  law,  with  Cy  as  the  applied  voltage  and 
Co  as  the  critical  voltage  in  the  equation.     It  should  then  follow 
the  quadratic  law  for  all  higher  voltages.     On  the  weathered 
conductors  used  in  practice,  the  quadratic  law  is  followed  over 
the  whole  range  of  voltage,  starting  at  eo. 

Example :— In  order  to  show  the  variation  in  corona  loss 
at  different  voltages  and  for  different  weather  conditions,  Table 
E  has  been  calculated  for  No.  o  stranded  copper  conductors 
(105560  circ.  mils,  0.373  in.  diameter)  and  for  steel  reinforced 
aluminum  conductors  (167800  circ.  mils,  0.501  in.  diameter) 
having  an  quivalent  resistance  but  greater  diameter.  F.  W. 
Peek's  formulas  were  used  and  the  following  assumptions  were 
made : — 

/        =60  cycles, 
mo     ^  0.87 
m,     =  0.72 
go      =  53-6 


CORONA  EFFECT 


37 


r 
s 
b 
t 
s 
r 


4 


=  0.186  in.  for  copper  =  0.250  in.  for  aluminum. 
=;  144  inches  (delta  arrangement  of  conductors). 
=  28.9  corresponding  to  an  altitude  of  1000  feet 
=  77  degrees  F.  S  therefore  =  0.967. 

=  774  for  copper  =  576  for  aluminum 
logio  774  =  2.89  and  logio  576  =  2.76 


iL=  0.036  for  copper  and  0.0415  for  aluminum. 
s 


DISRUPTIVE  CRITICAL  VOLTAGE— Fair  Weather 

eo  =  2.302  nto  gohr  log\a  —  {20) 

effective  kv  to  neutral 
For  the  Copper  Conductors 

(To  =  2.302  X  0.S7  X  53.6  X  0.967  X  0.186  X  2.89 
=  55.8  kv  to  neutral  (96500  volts  between  conductors). 
Table  XXII  gives,  by  interpolation,  the  limitation  of  e,  for 
above  conditions,  as  96500  volts  between  conductors.  To  find 
Co  to  neutral  for  any  other  altitude  or  temperatures  insert  the 
corresponding  values  of  8  for  the  altitude  and  temperature  in 
the  formula. 

TABLE  D— WORKING  TABLE- S  (DENSITY)  VALUES 

Altitude  and  Temperature  Correction  Factors 

17.9b 


8    = 


459  +  t 
t  =  temperature  in  degrees  F. 


where  b  =  barometric  pressure  in  inches  and 


.Mtitude   ill 

Barometer 

0    Values  for  Different  Temp. 

Feet 

III 

o»C. 

25°  c. 

see. 

Inches 

1 

(S:!"  F.) 

(77°  F.) 

(I3i»  F.) 

Sea  Level 

30.0 

76.2 

1.09 

* 
1. 00 

0.925 

500 

2945 

74.8 

1.07 

0.985 

0.910 

1000 

28.90 

73-3 

1.05 

0.967 

0.892 

1500 

28.30 

71.8 

1.03 

0.947 

0.873 

2000 

27,80 

70.7 

I.OI 

0.932 

0.860 

2500 

27.25 

69.2 

0.955 

0.912 

0.841 

3000 

26.80 

68.0 

0.980 

0.897 

0.827 

4000 

2575 

65.3 

0.940 

0.860 

0.793 

5000 

24.70 

62.7 

0.902 

0.827 

0.762 

6000 

23.90 

60.7 

0.875 

0.800 

0.738 

7000 

22.95 

58.3 

0.840 

0.770 

0.710 

8000 

22.05 

56.0 

0.805 

0.738 

0.682 

9000 

21.30 

54-1 

0.778 

0.712 

0.657 

10  000 

20.50 

52.1 

0.750 

0.687 

0.633 

12000 

19.00 

48.3 

0.697 

0.637 

0.588 

14000 

17.55 

447 

0.643 

0.588 

0.543 

15000 

16.90 

42.9 

0.618 

0.566 

0.522 

♦This  column  contains  the  values  for  8  which  were  used 
in  determining  the  values  of  Co  in  Table  XXII.  That  is,  the 
■values  for  sea  level  in  Table  XXII  multiplied  by  these  8  values 
gives  the  c,  values  of  the  table  for  the  higher  altitudes. 

For  the  Aluminum  Conductors 

eo  =  2.302  X  0.8/  X  33-6  X  0.967  X  0.23  X  2.76 
=  71.5  kv  to  neutral  (123500  volts  between  conductors). 
Table   XXII    gives    (by   interpolation)    the   limitation    for 
above  conditions  as  123500  volts  between  conductors. 

To  find  Co  to  neutral  for  any  other  altitude  or  temperature 
insert  the  corresponding  value  of  8    for  that  altitude  and  tem- 
perature in  the  formula. 
DISRUPTIVE  CRITICAL  VOLTAGE— Stormy  Weather 

Co     during  storm  =  approximately  80  percent  eo  during 
fair  weather. 

For  the  Copper  Conductors 

eo     for  storm  =  55.8  X  O.80  =  44.6  kv  to  neutral  or 
77000  volts  between  conductors. 

For  the  Aluminum  Conductors 

fo     for  storm  =  715  X  0.80  =  57.2  kv  to  neutral  or 
98800  volts  between  conductors. 

VISUAL  CRITICAL  VOLTAGE— Fair  Weather 

o./8g 


e,  =  2.302  Wv  goSr 
effective  kv  to  neutral 


(      0./S9  y,     s 


(^/) 


For  Copper  Conductors 


e,  -  i^o2Xo.72Xs3.6Xo.967Xou86(i+-^^\  2J9 

=  66.4  kv  to  neutral  (115  000  volts  between  conductors). 

To  find  e,  to  neutral  for  any  other  altitude  and  temperature, 
insert  the  corresponding  values  of  8  for  that  altitude  and  tem- 
perature in  the  formula  above. 

For  the  Aluminum  Conductors 

Cr  -  2^O2X0.72XS3-6Xo.<f67X0.2s(/+—-^\  2.76 

=  82  kv  to  neutral  (141S00  volts  between  conductors). 

To  find  e,  to  neutral  for  any  other  altitude  and  temperature, 
msert  the  corresponding  values  of   S  for  that  altitude  and  tem- 
Jerature  in  the  formula  above. 
POWER  LOSS  _ 

P=Y  ^  +  ^^^  J-T  ^^  "  ^°^'  "^ ^"^ 

kw  per  mile  of  each  conductor 

The  corona  power  loss  corresponding  to  various  conditions 
for  the  above  circuit  has  been  calculated  by  formulae  (22)  and 
(22A).  They  are  given  in  Table  E.  However,  in  order  to 
illustrate  the  application  of  the  power  loss  formula  the  losses 
for  the  following  conditions  are  determined  below.  Assuming 
that  the  No.  0  stranded  copper  conductors  will  be  operated  at 
105  kv  between  conductors  (60.7  kv  to  neutral). 

For  Fair  Weather— Max.  Temp.  50  degrees  C.  (122  degrees 
F.)— £„  =  51.3  kv. 


P  =  ^^U  +  ^5) 


AjL  {.e-  eoY-iar^ {ji2) 

kw  per  mile  of  each  conductor 


^  °  TSgT  e^  +  ^.^)  X  "•"3'^  (^-7  -  Sr.3)^  /o"' 

=  1.2  kw  per  mile  of  each  conductor  or  3.6  kw  per  mile 
for  three  conductors. 

For  Stormy  Weather— Max.  Temp.  25  degrees  C.  {77  de- 
grees F.)— £.  =  55.8  X  0.8  =  44.6  kv. 
3go 
^  =  ~^:^  (^  +  ^^)  X  °-°S^  (^-7  -  44.6Y  10-^  {22 A) 

=  3.2  kw  per  mile  of  each  conductor  or  9.6  kw  per  mile 
for  three  conductors. 

By  applying  formula  (20)  to  the  ^.bove  case  it  develops 
that  the  fair  weather  values  of  e,  are  for  25  degrees  C.  {77 
degrees  F.)  96500  kv  and  for  50  degrees  C.  (122  degrees  F.) 
B8800  kv  between  conductors.  Table  XXII  values  for  25  de- 
grees C.  {77  degrees  F.)  confirm  this. 

Table  E  values  for  corona  loss  indicate  that  No.  o 
copper  conductors  can,  with  144  inch  delta  arrangement 
of  conductors  and  looo  ft.  elevation  be  used  at  line  volt- 
ages as  high  as  100000  volts  without  excessive  corona 
loss  during  stormy  weadier.  At  100  000  volts  and  as- 
suming a  25  degrees  C.  {yy  degrees  F)  temperature  dur- 
ing fair  weather  and  storm  conditions,  the  corona  losses 
would  be  o.i  kw  per  mile  for  fair  weather  and  6.5  kw 
per  mile  for  stormy  weather.  If  the  transmission  is 
single  circuit  100  miles  long  and  without  branches,  has 
an  average  altitude  of  1000  feet  and  the  storm  condition 
existed  throughout  the  length  of  the  circuit,  the  power 
loss  due  to  corona  would  be  6.5  X  100  or  650  kw.  The 
capacity  of  such  a  circuit  at  100  000  volts  (see  Table 
XX)  would  be  roughly  15000  kw  at  ten  percent  l^R 
loss.     The  storm  corona  loss  therefore  would  represent 

650 
or  4.3  percent.     This,  in  addition  to  ten  percent 

PR  loss,  would  represent  approximately  14  percent  loss 
in  transmission  during  the  storm  conditions. 

In  the  above  case  it  would  probably  be  considered 
good  engineering  (so  far  as  corona  loss  is  concerned) 


38 


CORONA  EFFECT 


TABLE  XXM-APPROXIMATE  VOLTAGE   LIMITATIONS 

RESULTING   FROM  CORONA 


STRANDED    COPPER    CONDUCTORS 


AND 
MILS 

ii 

i  i 

CLEVATlO^ 
IN 

LIMIT  IN  KILOVOLTSIBETWEENI CONDUCTORS 
3  PHASE;  FOR  VARIOUS!  SPACINQSI       X 

AND 

OtHOULAB 

MILS 

SS 

ii 

ELEVATION 

IN 

FEET 

LIMIT  IN  KILOVQLTS  BETWEEN  CONDUCTORS 
3  PHASE  iFOFI  VARIOUS  SPACINQS      X 

3 
FT. 

4 

FT 

5 

FT. 

6 

FT. 

7 
F_T 

8 
FT. 

9 

FT 

II 

FT 

13 

FT 

16 

FT 

19 

FT 

25 

FT. 

3 
FT 

4 

FT 

5 
FT 

6 

FT 

7 
FT 

8 
FT. 

9 
FT 

II 
FT 

13 

FT 

15 

FT 

19 
FT 

26 
FT 

4 

.232 

BEA  LEVEL. 
lOOO 
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43 

tt 
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44 

73 

260  000 

676 

SEA  LEVEL 

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■  000 
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X  For  single  phase  or  2  phase  multiply  the  3  phase  values  by  1.16.    The  above  are  the  disruptive  critical  voltage  values  for  fair  weather  based  upon  a  tt;"»- 
perature  of  25°  C.  (77°  F.)  and  values  for  M,  of  0.S7  for  stranded  and  0.%  for  solid  conductors.     Derived  by  Peek's  formula:   Kilovolts  to  neutral  ^  2.302  M„  G„  S 

T  =  Temperature  in  degiees  P.;  B  =■ 


S 


R  I-"K  ,.  -j^.  where  G„  —  53.6  Kilovolts  per  inch;  S  =  Spacing  in  iiitlies;  R  =  Radius  of  conductor  in  inches;  S  - 


R 
Barometer  pressure  in  inches. 


i7.q  B 
459+T' 


CORONA  EFFECT 


39 


to  operate  the  No.  o  copper  conductors  at  as  high  a  line 
voltage  as  too  coo  volts.  If,  however,  for  other  reasons, 
ijoooo  is  selected  as  the  desirable  operating  voltage, 
then   either   a   large   diameter   copper   conductor   or   an 


aluminum  conductor  having  a  greater  diameter  hut  an 
equivalent  conductivity  to  that  of  the  No.  o  copper  con- 
ductor should  be  selected. 


TABLE  E— COMPARISON  OF  CORONA  LOSS 


For 

No.  0  Stranded  Copper  Conductors  105  S60cir.mil  (diameter  0.373  in.)  and  equivalent  Aluminum  Conductors  167  800 dr.  mil  (diametei  0.501  in.) 
Conductor  Spacing  (s)  Delta  -  144  in.    Altitude  1000  feet— Barometer  28.9  inches.    Calculated  from  formula  (22) 

Kilovolts 

Corona  Loss  in  Kw.  per  Mile  for  Three  Conductors  at  60  CycUs 

Fair  Weather- 

(Formula  22) 

Stormy  Weather— (Formula  22-A) 

Between 
Conduct- 
ors 

To 
Neutral 

No.  0  Copper 
Radius  0.186  in. 

Aluminum 
Radius  0.25  in. 

No.  0  Copper 
Radius  0.186  in. 

Alnroinuro 
Radius  0.2S  in. 

0°  C 
32°  F 
6-1. OS 
««-60.5 

25°  C 
77°  F 
5-0.967 
eo-5S.7 

50*  C 

122°  F 

5=0.892 

eo-51.3 

0°  C 

32°  P 

5=1.05 

eo=77.S 

25»C 

77°  F 

5-0.967 

eo-71.5 

SO*  c 

122*  F 

5-0.892 

eo-66.0 

0*  C 

32*  F 

5  =  1.05 

«o-48.4 

25*  C 

77°  F 

5-0.967 

eo-44.5 

SO*  C 

122"  F 

5-0.892 

eo-41.0 

0*  C 

32*  F 

S-\.OS 

(•-62. 

2S*  C 

77*  F 

5-0.967 

eo-S7.2 

SO*  C 

122*  F 

5-0.892 

<.-52  7 

100 
110 
120 

57.8 
63.5 
69.2 

0.0 
0.3 
2.6 

0.1 
2.3 
6.7 

0.2 

6.0 

12.8 

0 

0 
0 
0 

0 

0 

0.4 

0.3 

7.8 

14.8 

6.S 
13   3 
22   6 

11.3 
20.3 
32.0 

0 

0 

2.0 

0 

17 
6.2 

1.1 

4.6 

12.6 

130 
140 
150 

75.1 
80.8 
86.7 

7.25 
13.8 
22.4 

13.9 
23.3 
35. S 

22.6 
34.8 
50.2 

0.0 
0.3 
3.3 

0.5 

3.7 
9.9 

3.8 
10.1 
19.7 

24.4 
35. 8 
SO. 2 

34. 6 
48.7 
66. 

46.  S 
63.7 
84. 

6.7 
13  9 
24. 

13.7 
23  8 
37  2 

23.2 
36.4 
S3. 3 

160 
180 

92.4 
104.8 

35.0 
66.0 

49.8 
89.0 

67.7 
115.0 

8.7 
29.3 

18.7 
47.3 

32.2 
69.5 

66. 
108. 

85. 
135. 

106 
163. 

36. 

72. 

53. 
96. 

73. 
I2S. 

fl   +  25 

Note:    At  25  cycles  the  losses  would  be 

f  +  25 


25   +  25 


50 


■  times  the  above  table   values.   For  conductors  in   a   cow   (flat    spacing)  the 


60   +  25  85 

corona  loss  would  be  reduced  below  the  values  for  delta  or  triangular  arrangement.  For  the  higher  voltages  in  the  above  table  the  conductor  spacings  would,  in  aa 
actual  installation,  be  greater  than  144  in.  (upon  which  basis  the  table  values  are  given)  thus  giving  actually  less  corona  loss  for  the  higher  voltages  thao  indicated 
by   the   table   values. 


The  accompanying  photograph  illustrating  corona 
on  an  experimental  line  is  published  with  the  kind  per- 
mission of  F.  W.  Peek,  Jr. 

Since  the  formulas  pertaining  to  corona  effect  are  to 
some  extent  worked  up  from  test  data  they  may  be 
slightly  changed  from  time  to  time.  In  case  the  problem 
Jit  hand  seems  vitally  near  fhe  critical  point  it  will  be  well 
to  consult  the  latest  literature  at  that  time  as  an  additional 
check  on  the  work. 


roROX.A   .\T  230   Kv.    t.ig  cm.    dt.wietf.r,   0.47"   c.\ble,   310  cm. 

10  FEF.T  SPACING. 


CHAPTER  V 

SPEED  OF  ELECTRIC  PROPOGATION     RESONANCE 
PARALLELING  TRANSMISSION  CIRCUITS 
HEATING  OF  BARE  CONDUCTORS 


.   SPEED  OF  ELECTRIC  PROPAGATION 

ASTRONOMERS  and  investigators  by  various 
methods  of  determination  have  arrived  at 
sHghtly  different  values  for  the  speed  of  light. 
The  Smithsonian  Physical  Tables  give  i86  347  miles  per 
second  as  a  close  average  estimate.  In  electrical  engi- 
neering, the  speed  of  light  is  usually  stated  as  approxi- 
mately 3  X  io^°  centimeters  per  second.  This  is  the 
equivalent  of  186  451  miles  per  second.  The  speed  of 
electrical  propogation  (assuming  zero  losses)  is  the 
same  as  .that  of  light. 

ELECTRIC  WAVE  LENGTH 

Suppose  a  frequency  of  60  cycles  per  second  is  im- 
pressed upon  a  circuit  of  infinite  length.  At  the  end 
of  one  sixtieth  of  a  second  the  first  impulse  (neglecting 
retardation  due  to  losses)  will  have  traversed  a  distance 
of  186  347  -f-  60  or  3106  miles.  A  section  of  such  a  cir- 
cuit 3106  miles  long  would  be  designated  as  having  a 
full  wave  length  for  a  frequency  of  60  cycles  per 
second. 

In  Fig.  14,  the  dotted  line  or  one  cycle  wave  is 
shown  as  extending  over  a  circuit  3106  miles  long.  In 
this  case,  when  the  first  part  of  the  wave  arrives  at  a 
point  3106  miles  distant,  the  end  of  the  same  wave  is 
at  the  beginning  of  the  circuit.  For  each  half  wave 
length  the  current  is  of  equal  value  but  flowing  in  op- 
posite directions  in  the  conductor.  Such  a  circuit  is 
designated  as  of  full  wave  length.  Since  the  velocity  of 
the  electric  propagation  is  slightly  less  than  that  of  light, 
being  slightly  retarded  due  to  resistance  and  leakage 
losses,  the  actual  wave  length  will  be  slightly  less  than 
3106  miles.  Thus  for  a  300  mile,  60  cycle,  three-phase 
circuit  consisting  of  No.  000  copper  conductors  having 
10  ft.  flat  spacing,  the  wave  length  is  calculated  to  be 
2959  miles.  The  wave  length  of  such  a  circuit  is  in- 
dicated by  the  heavy  line  on  the  accompanying  sketch. 
In  the  case  of  this  particular  circuit  the  electric  field  has 
been  retarded  approximately  five  percent,  due  to  the 
losses  of  the  circuit,  as  indicated  by  the  displacement  of 
the  dotted  and  full  line  curves. 

QUARTER   WAVE  RESONANCE 

If  the  end  of  a  long  trough  filled  with  water  is 
struck  by  a  hammer,  the  impact  will  cause  a  wave  in  the 
water  to  start  in  front  of  the  point  of  impact  and  travel 
to  the  far  end  of  the  tank.  When  this  wave  reaches  the 
far  end  of  the  tank  it  will  be  reflected,  traveling  back 
toward  the  point  of  origin,  but  on  account  of  resistance 
encountered  it  will  be  of  diminishing  height  or  ampli- 
tude. If,  at  the  instant  it  gets  back  to  the  point  of  origin, 
the  end  of  the  tank  is  again  struck  by  the  hammer,  the 


resulting  impulse  will  be  that  due  to  the  second  hammer 
blow  plus  that  remaining  from  the  first  blow.  The  re- 
sult will  be  that  the  second  wave  from  the  near  end  of 
the  tank  will  be  of  greater  amplitude  than  the  first  wave. 
If  when  the  second  wave  arrives  back  at  the  near  end, 
the  end  of  the  tank  is  struck  again  with  the  hammer  the 
resulting  third  impulse  will  be  of  greater  amplitude  than 
the  second  impulse.  If  at  the  instant  of  the  return  of 
each  succeeding  impulse  the  end  of  the  tank  is  struck, 
the  result  will  be  cumulative  and  each  succeeding  wave 
will  be  of  greater  magnitude  than  the  one  preceeding 
until  the  point  is  reached  where  the  losses  due  to  resist- 
ance become  sufficient  to  prevent  a  further  increase  in 
amplitude  of  the  wave. 

Under  certain  conditions  a  similar  phenomenon  may 
occur  in  electric  circuits  and  this  is  known  as  "quarter 
wave  resonance".     If  an  electric  impulse*  is  sent  into  a 


1.— WAVE  LENGTH  lE    LOSSES  WERE  NOT  PRESENT  TO  RETARD  TmE   ELECTRIC  PROPAGATION  =  3  106  MILES — »* 

FIG.    14 — WAVE  LENGTH   OF  6o  CYCLE  CIRCUIT 

conductor,  such  as  a  transmission  circuit,  this  impulse 
travels  along  the  conductor  at  the  velocity  of  light.  If 
the  circuit  is  open  at  the  other  end,  the  impulse  is  there 
reflected  and  returns  at  the  same  velocity.  If  at  the 
moment  when  the  impulse  arrives  at  the  starting  point 
a  second  impulse  is  sent  into  the  circuit,  the  returned 
first  impulse  adds  itself  to,  and  so  increases  the  second 
impulse;  the  return  of  this  second  impulse  adds  itself  to 
the  third  impulse,  and  so  on ;  that  is,  if  alternating  im- 
pulses succeed  each  other  at  intervals  equal  to  the  lime 
required  by  an  impulse  to  travel  over  the  circuit  and 
back,  the  effects  of  successive  impulses  add  themselves, 
and  large  currents  and  high  e.m.f.'s  may  be  produced 
by  small  impulses.  This  condition  is  known  as  quarter 
wave  electric  resonance.  To  produce  this  condition,  it 
is  necessary  that  the  alternating  impulses  occur  at  time 
intervals  equal  to  the  time  required  for  the  impulses  to 
travel  the  length  of  the  line  and  back.  For  example, 
the  time  of  one  half  wave  or  cycle  of  impressed  e.m.f. 


*For  a  complete  study  of  this  subject  see  "Transient  Electric 
Phenomena  and  Oscillations"  by  C.  P.  Steinmetz,  from  which 
the  above  description  of  quarter  wave  resonance  has  largely 
been  taken. 


PROP  AG  A  TION— RESONANCE— PARALLELING    CIRCUITS 


41 


is  the  time  required  by  light  to  travel  twice  the  length 
of  the  line,  or  the  time  of  one  complete  cycle  is  the  time 
light  requires  to  travel  four  times  the  length  of  the  line. 
Stated  another  way,  the  number  of  cycles  or  frequency 
of  the  impressed  alternating  em.f.'s  in  resonance  condi- 
tion, is  the  velocity  of  light  divided  by  four  times  the 
length  of  the  line;  or  to  have  free  oscillation  or  reson- 
ance condition,  the  length  of  the  line  is  one  quarter 
wave  length  of  light.  The  cycles  at  which  this  condi- 
tion is  reached  (if  there  were  no  losses  present)  would 
be  determined  as  follows: — 

Pre  uen      ———^^^l— 

'         ■'         Length  in  miles    ^   '' 


*  Length  in  miles  ^— f 

Frequency 


('4) 


RESONANCE  LENGTHS  OF  CIRCUITS 


Commercial  frequencies  are  so  low  that  to  reach  a 
quarter  wave  resonance  condition  with  them  the  circuit 
would  have  to  be  of  great  length.  The  following  values, 
for  the  sake  of  simplicity,  are  based  upon  the  assump- 
tion that  there  are  no  losses  in  the  circuit. 

Fundamental  Frequency      Resonance  Length  Wave  Length 

IS  c>xlcs   3106  miles  12434  miles 

25  cycles  1863  miles  7452  miles 

40  cycles   1165  miles  4660  miles 

60  cycles  776  miles  3106  miles 

The  above  lengths  are  based  upon  the  impressed  or 
fundamental  frequencies.  If  these  impressed  fre- 
quencies contain  appreciable  higher  harmonics,  some  of 
the  latter  may  approach  resonance  frequency  and,  if  of 
sufficient  magnitude,  may  cause  trouble.  Thus  the 
length  of  circuit  corresponding  to  resonance  conditions 
of  various  harmonics  of  the  fundamental  is  given  below. 


Cycles 

Harmonics                                | 

3rd. 

5th. 

7th. 

15 
25 
40 
60 

1035  miles 
621  miles 
388  miles 
258  miles 

631  miles 
372  miles 
233  miles 
155  miles 

444"  miles 
266  miles 
166  miles 
III  miles 

Thus  an  impressed  frequency  of  60  cycles  will  not 
produce  quarter  wave  electric  resonance  unless  the  cir- 
cuit be  approximately  776  miles  long.  If  a  third  har- 
monic, however,  is  present  in  the  impressed  wave,  this 
harmonic  will  develop  quarter  wave  resonance  in  a  cir- 
cuit approximately  258  miles  long,  a  5th  harmonic  in  a 
circuit  approximately  155  miles  long,  and  a  7th  har- 
monic in  a  circuit  approximately  iii  miles  long. 

The  above  values  are  based  upon  no  losses  being 
encountered  in  transmission.  Obviously  this  is  an  in- 
correct assumption,  as  electric  propagation  is  always 
accompanied  by  more  or  less  loss,  depending  upon  the 
fundamental  constants  (resistance  and  leakage)  of  the 
circuit.  The  effect  of  such  losses  is  to  retard  the  ve- 
locity of  the  electric  propagation,  usually  by  an  amount 
of  five  to  ten  percent  below  that  of  light.  The  above 
values  of  circuit  lengths  representing  a  condition  for  re- 
sonance may  therefore  be  as  much  as  ten  percent  above 
the  actual  lengths. 

An  investigation  of  the  effects  of  higher  harmonics 


of  the  impressed  wave  is  of  importance  in  connection 
with  very  long  distance  transmission  systems. 

PARALLELING  TRANSMISSION  CIRCUITS 

Transmission  lines  are  frequently  constructed  with 
duplicate  circuits  which  are  normally  operated  in  paral- 
lel. In  other  cases  two  circuits  may  lead  from  the  gen- 
erating station  in  divergent  directions  and  at  some  dis- 
tant point  come  together  and  be  connected  in  parallel. 

If  the  two  circuits  are  fed  from  different  genera- 
tors, or  sources  of  supply,  the  only  condition  necessary 
for  paralleling  the  circuits  is  that  the  phase  rotation  of 
the  two  circuits  be  the  same  and  that  the  regulation  in 
speed  of  the  prime  movers  of  the  generators  feeding 
the  two  systems  can  be  adjusted  so  as  to  bring  the 
phases  of  the  two  circuits  together  for  paralleling. 

If,  however,  the  two  circuits  which  are  to  be  con- 
nected in  parallel  are  fed  from  the  same  source  of  sup- 
ply, the  case  may  become  involved.  There  will  be  no 
trouble  in  obtaining  the  correct  phase  rotation,  for 
should  the  circuits  not  rotate  alike,  it  is  only  necessary 
to  transpose  any  two  of  the  connections  of  either  of  the 
circuits  (assuming  that  the  circuits  are  three-phase). 
The  other  condition  to  be  met  is  that  the  phases  of  both 
circuits  to  be  paralleled  are  the  same,  i.  e.,  the  volt- 
ages in  the  phases  to  be  paralleled  must  pass  through 
their  zero  and  maximum  values  at  the  same  instant. 

If  neither  circuit  has  transformers  between  the 
points  where  they  are  to  be  connected  in  parallel,  their 
phases  will  coincide  and  there  will  be  no  trouble  about 
connecting  them  in  parallel.  If  one  circuit  has  no 
transformers  and  the  other  has  transformers,  the  phase 
relations  of  the  two  circuits  will  depend  upon  the  kind 
of  transformer  connections  employed.  Suppose  it  is 
assumed  that  the  raising  transformers  are  connected 
delta  to  star  and  the  lowering  transformers  are  con- 
nected delta  to  delta.  With  these  connections  the 
phases  of  the  two  circuits  will  be  30  electrical  degrees 
apart  and  it  will  be  impossible  to  parallel  the  circuits. 
In  other  words  one  delta-star  or  star-delta  transformer 
connection  produces  a  phase  displacement  of  30  degrees. 
It  will  be  obvious  that  a  second  delta-star  or  star-delta 
connection  will  restore  the  original  phase  relation.  A 
delta-delta  connection  or  a  star-star  connection  does 
not  affect  the  phase  relations.  If  both  circuits  have  an 
even  number  of  star  and  even  number  of  delta  wind- 
ings, the  equivalent  resultant  will  be  the  same  as  if  all 
the  connections  were  either  delta-delta  or  star-star; 
hence,  there  will  be  no  resultant  change  in  phase  rela- 
tions and  the  two  circuits  can  be  paralleled  with  each 
other  or  with  a  circuit  having  no  transformations.  If, 
however,  both  circuits  have  an  odd  number  of  delta  and 
an  odd  number  of  star  windings,  any  attempt  to  re- 
solve them  into  the  equivalent  number  of  delta-delta 
and  star-star  connections  will  leave  one  star  and  one 
delta;  the  effect  is  the  same  as  if  there  was  one  star- 
delta  connection  in  the  circuits.  This  will  twist  the 
phase  relations  of  the  terminals  30  degrees  out  of  phase 
from  the  generators.     Since  both  circuits  will  have  an 


42 


PROPAGATION— RESONANCE— PARALLELING    CIRCUITS 


equivalent  phase  displacement,  they  can  be  paralleled 
with  one  another,  but  since  both  are  30  degrees  out  of 
phase  with  the  generators,  they  cannot  be  paralleled 
with  a  line  having  no  transformations;  nor  with  a  line 
having  an  even  number  of  star  and  delta  connections. 

When  the  phase  angles  of  the  two  transmission  cir- 
cuits (receiving  their  power  from  a  common  source)  are 
known  to  be  such  as  to  permit  of  parallel  operation  it 
is  then  necessary  to  phase  them  out  before  connecting 
the  circuits  together.  The  phase  rotation  can  be  checked 
most  readily  by  means  of  a  polyphase  motor  connected 
first  to  one  circuit  and  then  to  the  other,  being  careful 
to  connect  the  leads  in  the  same  order  in  each  case.  If 
the  motor  runs  in  the  same  direction  from  both  circuits, 
the  phase  rotation  of  the  circuits  will  be  the  same.  The 
phase  angle  can  be  readily  tested  by  means  of  a  single- 
phase  synchroscope*.  In  case  a  polyphase  motor  and 
synchroscope  are  not  available,  the  phasing  out  of  the 
circuits  may  be  accomplished  by  the  use  of  a  voltmeter 
and  transformer.**  As  an  illustration,  assume  that 
from  a  4400  volt  bus  in  a  generating  station  a  4400  volt 
transmission  circuit  extends  for  some  distance  from  the 
station.  A  second  transmission  circuit  fed  from  the 
same  bus  but  containing  both  raising  and  lowering 
transformers  is  to  be  paralleled  at  the  farther  end  with 
the  4400  volt  circuit  which  contains  no  transformers. 
The  phase  angles  of  the  lines  are  assumed  to  be  such 
as  to  permit  paralleling  the  two  circuits,  with  proper 
connections. 

One  of  the  transmission  circuits  is  connected  to 
one  side  of  the  paralleling  switch  as  in  Fig.  15  and  the 
other  circuit  to  the  other  side  of  the  same  switch.  The 
three  terminals  on  one  side  of  the  switch  may  be  tagged 
1-2-^.  Likewise  the  three  terminals  on  the  other  side 
of  the  switch  may  be  tagged  4-5-6.  Connect  any  two 
terminals  together  (i  and  4  in  this  case)  by  a  jumper. 
Take  voltage  readings  across  the  corresponding  ter- 
minals 2  to  5,  s  to  6,  and  j  to  5,  ^  to  6.  From  these 
voltage  readings  it  is  a  simple  matter  to  indicate  by  a 
vector  diagram  the  relative  phase  relations  at  the  switch 
contacts  of  the  two  circuits  to  be  paralleled.  In  the  case 
illustrated,  the  readings  indicate  that  the  relative  voltage 
relations  on  the  two  sides  of  the  paralleling  switches  are 
as  indicated  by  the  full  line  delta  1-2-^,  and  the  broken 
line  delta  4-5-6.  It  will  be  seen  that  phase  j-j  will 
parallel  with  phase  4-5,  that  phase  1-2  will  parallel  with 
phase  6-5  and  phase  .?-j  will  parallel  with  phase  4-6. 
In  order  to  bring  about  this  phase  relation  it  will  be  nec- 
essary to  change  the  transformer  connections  on  the 
low-tension  side  of  the  lowering  transformers,  inside 
of  the  delta.  That  is  the  6  end  of  the  transformer 
windings  5-6  will  be  connected  to  the  4  end  of  transfor- 


*These  tests  are  described  in  an  article  on  "Phasing  Out 
High  Tension  Lines"  by  E.  C.  Stone  in  the  Journal  for  Nov. 
1917,  p.  448- 

**This  method  is  described  in  an  article  on  "Determination 
of  Polarity  of  Transformers  for  Parallel  Operation"  by  W. 
M.  McConahey,  in  the  Journal  for  July  1912,  p.  613.  See  also 
article  on  "Polarity  of  Transformers"  by  W.  M.  Dann  in  the 
Journal  for  July  1916,  p.  350. 


mer  4-5.  The  4  end  of  transformer  4-6  will  be 
connected  to  the  5  end  of  transformer,  5-6  and  the  6  end 
of  transformer  4-6  will  be  connected  to  the  5  end  of 
transformer  4-5.  These  changes  will  shift  the  position 
of  the  delta  4-5-6  so  that  it  will  coincide  with  delta 
1-2-3.  -^  further  test  of  voltage  between  switch  term- 
inals 2X.0  5  and  5  to  d  should  indicate  zero  voltage  across 
the  switch  terminals  to  be  connected  together,  in  which 
case  the  paralleling  switches  may  be  closed.  In  order  to 
measure  the  voltage  across  the  paralleling  switch  con- 
tacts it  will  usually  be  necessary  to  employ  a  potential 
transformer.  This  transformer  and  voltmeter  should  be 
capable  of  withstanding  1.73  times  the  voltage  of  the 
circuit  for,  with  the  connections  given  in  Fig.  15,  one 
reading  gave  7610  volts,  whereas  the  voltage  of  the  cir- 
cuit was  only  4400  volts. 

In  case  there  is  a  ground  on  both  systems,  the  plac- 
ing of  a  jumper  across  two  of  the  switch  contacts  would 
result  in  a  short-circuit.  This  jumper  should  not  be 
placed  across  the  switch  until  it  b.as  been  shown  by  con- 
necting a  transformer  across  these  two  contacts  that  no 
potential  exists  between  them. 

HEATING  OF  BARE  CONDUCTORS  IN  AIR 

If  the  circuit  is  long,  the  voltage  will  probably  be 
high   and   consequently   the   current   to   be  transmitted 


-4400  VOLT  LINE  DIRECT 
FROM  STATION  BUS 


!< 


2  TO  &-4400  VOLTS 


3  TO  6-4400  VOLTS 

SWITCH       ^  2  TO  6-7610  VOLTS 

C3)       (^    3  TO  6-ZERO  VOLTS 


FIG.    15 — TEST    FOR    PHASE    SEQUENCE 

small.  In  this  case,  the  heating  effect  of  the  current 
will  be  small  and  unimportant.  If,  however,  the  circuit 
is  short  and  an  unusually  large  amount  of  power  is  to  be 
transmitted,  the  current  will  be  large.  Since  the  PR 
loss  varies  as  the  square  of  the  current  and  directly  as 
the  resistance,  the  heat  generated,  if  the  current  is  large, 
may  be  sufficient  to  overheat  or  anneal  the  material  of 
the  conductors.  In  some  cases  of  unusually  large 
amounts  of  power  being  transmitted  short  distances,  the 
heating  effect  of  the  currents  resulting  may  be  sufficient 
to  limit  the  amount  of  power  that  can  be  transmitted 
at  a  given  voltage. 

Table  XXIII  should  be  consulted  in  cases  where 
the  circuit  is  short  and  the  amount  of  power  to  be  trans- 
mitted large.  In  this  table  are  columns  containing  cur- 
rent values  which  have  been  calculated  corresponding 
to  10,  25  and  40  degrees  C.  rise  in  temperature  for  vari- 
ous sizes  of  bare  copper  conductors  suspended  in  still 
air  at  a  temperature  of  25  degrees  C.  In  other  words 
these  current  values  are  based  upon  absolute  temper- 
atures of  35,  50  and  65  degrees  C.  The  current  values 
corresponding  to  a  temperature  rise  of  40  degrees  C. 


1 

i 


ERRATUM 

The  formula  used  in  calculating  the  values  for  table  XXIII,  page  43, 
embodied  the  only  available  information  on  this  subject  at  the  time  the  values 
were  calculated.  Recent  exhaustive  and  carefully  conducted  tests,  made  by 
George  E.  Luke,  indicate  a  wide  difTerence  in  results  from  the  table  values, 
especially  in  the  larger  size  conductors.  The  table  values  corresponding  to  40°  C 
rise  should  not,  therefore,  be  used. 

In  the  April,  1923  issue  of  the  Electric  Journal,  page  127,  appears  an  article 
entitled  "Current  Capacity  of  Wires  and  Coils"  in  which  Mr.  Luke  gives  the  results 
of  his  tests  and  the  empirical  formula  he  developed  as  a  result  of  the  test. 


0  'Oi*  o)  -, 


MUTAHfla 


.ijj 


PROP  AG  A  TION-RF.SONAXCE-PA  HA  U.Iil.lSG    CIRCUl  TS 

TABLE  XXIII-HEATING  CAPACITY  FOR  40o  C.  RISE 

OF   BARE   COPPER   CONDUCTORS   SUSPENDED  OUT  OF   DOORS 


43 


CONDUCTORS 

AMPERES- 

BARE 

APPROXIMATE  CARRYING  CAPACITY  IN 
TO  A   TEMPERATURE  RISF  DP  /in  r  (R/iOcn  non 

K  V  A    3  PHASE  corresponding] 

> 

d 

z 

CO 
CD 

AREA 

IN 

CIRCULAR 

MILS 

a.  CO 

1-  X 

u 

STILL   AIR   FOR 
TEMPERATURE 
RISES  STATED 

.~'  '^       .-...r  >_.>«,  urvt   nioo   ur-    nu   o  (bAbtU  UPO™    MlwrtKtO  IIM  UULUMN  MARKED 

FOR  40  C  RISE  )  FOR  BARE  COPPER  CONDUCTORS  SUSPENDED  IN   STILL  AlH 
OUT  OF  DOORS. 

220 
VOLTS 

440 
VOLTS 

550 
VOLTS 

MOO 
VOLTS 

2200 
VOLTS 

4000 
VOLTS 

4400 
VOLTS 

6000 
VOLTS 

6600 
VOLTS 

6900 
VOLTS 

FOR 
IO°C 
RISE 

FOR 
25°  C 
RISE 

f6r 

40°C 
RISE 

K.V  A. 

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KVA. 

KVA 

KVA 

KVA 

KVA 

KVA 

KVA 

KVA 

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1980 
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328C 
302C 
292c 

4050 
3760 
3600 

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3  080 
2  860 
2  740 

3  850 
3  580 

3420 

7  700 

7  ISO 
6  SSO 

'5  400 
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13  TOO 

2  T  900 
26  000 
24  800 

JO  800 

2  8  6O0 
27400 

42  000 
39  100 
37400 

46200 
42  800 

48  300 
44  800 
4  3  000 

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2  /   400 

26200 
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63  000 
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Formula  (from  Foster's  Hand  Book)  Amperes  =  r/M-yj  — -  for  stranded  conductors,  and  Amfirres-tiso-^  -rr-  for  solid  conductors,  where  7"=-  trmperatare 

rise  is  degrees  C,  D  ^  diameter  of  conductor  in  inches,  and  ^  =  resistance  in  ohms  per  mil  foot  at  the  final  lcmp«ralure,     Basetl  on  an  air  temperature  of  as 
degrees  C. 


44 


PROPAGATION— RESONANCE—PARALLELING    CIRCUITS 


(absolute  temperature  of  65  degrees  C.)  have  also  been 
expressed  in  the  form  of  k.v.a.,  three-phase  values  cor- 
responding to  various  transmission  voltages.  Thus  No. 
0000  stranded  bare  copper  conductors  suspended  in  still 
air  out  doors  at  25  degrees  C.  wrill  carry  750  amperes 
with  a  temperature  rise  of  40  degrees  C.  (absolute  tem- 
perature 65  degrees  C).  If  the  transmission  voltage  is 
220  volts,  the  corresponding  k.v.a.  value  will  be  285 
k.v.a.  three-phase  and  if  the  transmission  voltage  is 
10  000  volts,  13000  k.v.a.  may  be  transmitted  with  the 
same  temperature  rise. 


As  indicated  by  foot  notes  the  values  of  the  table 
were  calculated  by  formulas  from  Foster's  Handbook 
as  follows: — 


Amperes  =  Jioo -^ — -— for  stranded  conductor. 


.(<5) 


Amperes  =  1250 . 


-for  solid  conductor {26) 


Where 

T  =  Temperature  rise  in  degrees  C. 
D  =  Diameter  of  conductors  in  inches. 
R  =  Resistance   of   conductors   in   ohms   per   mil-foot   at 
final  temperature. 


CHAPTER  VI 


DETERMINATION  OF  FREQUENCY  &  VOLTAGE 


FREQUENCY  DETERMINATION 

Cost  of  Transformers — Sixty  cycle  transformers 
cost  approximately  30  to  40  percent  less  than  25  cycle 
transformers ;  or  stated  another  way,  25  cycle  trans- 
formers cost  approximately  40  to  66  percent  more  than 
60  cycle  transformers.  The  saving  in  first  cost  may 
vary  between  $1.50  and  $2.50  per  kv-a.  in  favor  of 
60  cycles.  Assuming  that  the  total  kv-a.  of  trans- 
former capacity  connected  to  a  transmission  circuit  is 
2.5  times  the  kv-a.  transmitted  over  the  circuit,  the  sav- 
ing in  favor  of  60  cycle  transformers  would  be  $3.75  to 
$6.25  or  an  average  of  $5.00  per  kv-a.  transmitted.  As- 
suming 20000  kv-a.  to  be  transmitted,  the  saving  in 
cost  at  $5.00  per  kv-a.  will  be  $100000  in  favor  of  60 
cycle  transformers.  The  actual  difference  in  cost  will 
depend  upon  the  type  of  the  transformers,  that  is, 
whether  water  or  self-cooled  and  also  upon  their  aver- 
age capacity.  The  difference  in  cost  will  be  greater  for 
the  self-cooled  type  and  for  the  smaller  capacities. 

Weight  and  Space  of  Transformers — The  less 
weight  of  60  cycle  transformers  makes  them  easier  to 
handle  and  they  require  less  space  for  installation. 

Higher  Reactance — Inductive  reactance  at  60 
cycles  is  2.4  times  its  value  at  25  cycles.  This  tends 
to  produce  poorer  voltage  regulation  of  the  circuit. 
Higher  reactance  has  one  advantage  for  the  larger  sys- 
tems in  that  it  tends  to  limit  short-circuit  currents  and 
thus  assists  the  circuit  opening  devices  to  function 
properly.  By  virtue  of  the  higher  reactance  it  might 
be  possible  in  some  cases  to  obtain  sufficient  reactance 
in  the  transformers  without  the  addition  of  current 
limiting  reactance  coils. 

Efficiency — The  efficiency  of  60  cycle  transformers 
is  usually  0.25  to  0.50  percent  higher  than  for  25  cycle 
transformers. 

Charging  Current — At  25  cycles  both  the  charging 
current  and  the  reactance  are  approximately  42  percent 
of  their  values  for  60  cycles.  This  tends  to  give  better 
regulation  and  usually  higher  efficiency  in  transmission. 
On  the  other  hand,  the  higher  transmission  efficiency 
may  be  offset  by  the  slightly  lower  eflficiency  of  25  cycle 
transformers.  In  cases  of  very  long  circuits  (par- 
ticularly if  the  circuits  are  in  duplicate  and  both  in  ser- 
vice) or  of  transmission  systems  embracing  many  miles 
of  high  tension  mains  and  feeders,  the  charging  currents 
may  be  so  great  as  to  limit  the  choice  in  transmission 
voltage.  On  the  other  hand  large  charging  currents 
may  be  permitted,  provided  under  excited  synchronous 
motors  are  used  at  various  parts  of  the  transmission 


system  for  partially  neutralizing  this  charging  current 
and  for  maintaining  con.jtant  voltage. 

Inductive  Disturbances— Lighimng,  switching  and 
other  phenomena  cause  disturbances  on  conductors  of 
transmission  circuits.  The  frequency  of  these  disturb- 
ances is  independent  of  that  impressed  on  the  system. 
After  the  removal  of  the  disturbing  influence  they 
oscillate  with  the  natural  frequency  of  the  line. 

The  natural  frequency  of  the  line  is  far  above  com- 
mercial frequencies  but,  if  the  transmission  line  is  long, 
there  may  be  some  odd  harmonic  present  in  the  funda- 
mental impressed  frequency  which  corresponds  with  the 
natural  period  of  the  line.  This  might  tend  to  produce 
an  unstable  condition  or  resonance.  This  condition  is 
somewhat  less  likely  to  occur  at  25  cycles. 

Summary — Although  there  are  a  number  of  large 
25  cycle  transmission  systems  in  operation,  they  were 
mostly  installed  before  the  design  of  60  cycle  converting 
apparatus  and  electric  light  systems  had  reached  their 
present  state  of  perfection.  Unless  it  is  desirable  to 
parallel  with  an  existing  25  cycle  system  located  in  ad- 
joining territory  without  the  introduction  of  frequency 
changers,  it  is  now  quite  general  practice  to  choose  the 
frequency  of  60  cycles.* 

VOLTAGE  DETERMINATION 

From  a  purely  economic  consideration  of  the  con- 
ductors themselves,  Kelvin's  law  for  determining  the 
most  economical  size  of  conductors  would  apply.  Kel- 
vin's law  may  be  expressed  as  follows : — 

"The  most  economical  section  of  a  conductor  is  that 
which  makes  the  annual  cost  of  the  PR  losses  equal  to 
the  annual  interest  on  the  capital  cost  of  the  conducting 
material,  plus  the  necessary  annual  allowance  for  depre- 
ciation". That  is,  the  economical  size  of  conductor  for 
a  given  transmission  will  depend  upon  the  cost  of  the 
conducting  material  and  the  cost  of  power  wasted  in 
transmission  losses.  The  law  of  maximum  economy 
may  be  stated  as  follows : — "The  annual  cost  of  the  en- 
ergy wasted  per  mile  of  the  transmission  circuit  added 
to  the  annual  allowance  per  mile  for  depreciation  and 
interest  on  first  cost,  shall  be  a  minimum". 

Attempts  have  been  made  to  determine  by  mathe- 
matical expression  the  most  economical  transmission 
voltage,  all  factors  having  been  taken  into  account. 
There  are  so  many  diverse  factors  entering  into  such  a 


♦For  a  complete  discussion  of  this  subject  see  a  paper  hy 
D.  B.  Rushmore  before  the  Schenectady  section  A.  I.  E.  E., 
May  17,  IQI2,  on  "Frequency"  and  an  article  by  B.  G.  Lamme 
on  "The  Technical  Story  of  the  Frequencies"  in  the  Journal. 
for  June,  1918,  p.  23a 


46 


DETERMINATION  OF  EREQUENCY  AND  VOLTAGE 


treatment  as  to  make  such  an  expression  complicated, 
difficult  and  unsatisfactory.  There  are  many  points  re- 
quiring careful  investigation,  not  embraced  by  Kelvin's 
law,  before  the  proper  transmission  voltage  can  be  de- 
termined.    Some  of  these  points  are  given  below. 

Cost  of  Conductors — For  a  given  percentage  energy 
loss  in  transmission,  the  cross-section  and  consequently 
the  weight  of  conductors  required  by  the  lower  and 
medium  voltage  lines  (up  to  approximately  30000 
volts)  to  transmit  a  given  block  of  power  varies  in- 
versely as  the  square  of  the  transmission  voltage.  Thus 
if  this  voltage  is  doubled,  the  weight  of  the  conductors 
will  be  reduced  to  one  fourth  with  approximately  a  cor- 
responding reduction  in  their  cost.  This  saving  in  con- 
ducting material  for  a  given  energy  loss  in  transmission 
becomes  less  as  the  higher  voltages  are  reached,  becom- 

TABLE    E  1— WEIGHT  OF  BARE  COPPER 
CONDUCTORS 


b 

z 
CO 

ED 

AREA  IN 

CIRCULAR 

MILS 

WEIGHT  IN  POUNDS 

PER  1000  FEET 
OF  CIRCUIT 

PER  MILE 
OF  CIRCUIT 

NUMBER  OF 
CONDUCTORS 

NUMBER  OF 
CONDUCTORS 

ONE 

TWO 

THREE 

ONE 

TWO 

THREE 

2  000  000 
/   900  000 
/   800  000 

i,  I<i0 
3  8  70 
SS(,n 

/i  340 
(/  740 
1  1   1  20 

19  S40 
17  <olO 
Ik  6S0 

32  630 
Z0  994 
293S7 

hS  2hQ 
bl  9S8 
Si7l4 

<?7  ?90 
92  9S2 
88  071 

/   700  000 

/  feoo  000 
/  soo  000 

S2S0 
4  630 

10  SOO 
<?880 
9  2h0 

IS  7S0 
I4S20 
I3S90 

27720 
26083 
2444t 

55  440 
52  lit 
48  892 

83'60 
78  249 
7?338 

1  -»oo  000 
1  zoo  000 
1  zoo  000 

4  320 
■4  010 
3  7/0 

S6-I0 
8020 
7420 

12  9&0 
12  030 
//   130 

22810 
21  1  73 
I9SS9 

451,20 
4234h 
39  I7S 

iS'430 
635/9 
58  767. 

1  100  000 

1  000  000 
9S0 000 

3  400 
3  090 
2  9-30 

6  SOO 
6  l%0 

ssto 

10  200 
9  270 
8  790 

17  9S2 
liSIS 
15470 

35  904 
32  630 
30940 

53856 
4  8  945 
4i4IO 

900 000 
eso  000 
eoo 000 

2  780 
2  <,20 
2  470 

ssto 

S240 

4  940 

?3-#0 
7  860 
7410 

14  67S 
13  83-) 
13  047 

793Si, 
27668 
26084 

44  034 
4/  502 
39  /26 

7  50  000 
TOO  000 

eso  000 

2320 
2lt0 
2010 

4h40 
4320 
.4020 

6  9i0 
6480 
(,0^0 

12  2S0 
II  40S 
10  6/3 

24  -500 
228/0 
2/226 

36  750 
342'X 
31  S39 

(,00  000 
sso 000 
soo  000 

1  SSO 
1  TOO 
1  S40 

3  700 
3400 
3  080 

SiiO 
S  100 
4&20 

9768 
8976 
8  131 

/9336 
17952 

itztz 

29304 
26928 
Z4393 

4S0 000 
400  000 
3  50  000 

1  390 
/  240 
1  O80 

2780 
24S0 
2  ItO 

4  1  70 
3  72  0 
3  240 

7339 
iS47 
S  702 

14  67? 
13094 

11404 

2201  7 
19  641 
17  lOi 

0000 

300  000 
2SO  000 
2/2  000 

?26 
77Z 
HSZ 

ISSZ 
1  S44 

1  306 

2778 

23l(, 
1  9  59 

4  889 
^076 
344S 

977S' 
8/52 
6S9<= 

/4  6  67 
12  228 
10  344 

000 
00 
0 

/68  000 
/33  000 
/06  000 

SIS 
4  II 
32(, 

1036 
822 
(,SZ 

1  SS4 

1  233 
978 

2  73S 
2  170 
1721 

5470 
4  340 
3442 

8205 
65/0 
5/63 

/ 
2 

3 

83  700 

a  400 
^3  600 

2S8 
20s 
/i3 

Sli 
4  10 
32t, 

774 
US 
4S9 

I  362 

/082 

86' 

2724 
2  /64 
/  72  2 

40?i, 
3  246 
2.SS3 

■4  1    700 
33  100 
26300 

129 

102 

SI 

2SS 

204 
It2 

387, 
30i 
243 

68/ 
^39 
-♦28 

/362 

I07S 

SSi 

2043 
/6/7 
/2  84 

I 

30  SOO 
lb  SOO 

i1 

IZS 
101 

19-2 
/S3 

338 
26? 

6  76 
S3% 

/  014 
807 

ing  increasingly  less  as  voltages  go  higher.  This  is  for 
the  reason  that  for  the  higher  voltages  at  least  two  other 
sources  of  losses,  leakage  over  insulators  and  the  escape 
of  energy  through  the  air  between  the  conductors 
(known  as  "corona")  appear.  In  addition  to  these  two 
losses,  the  charging  current,  which  increases  as  the 
transmission  voltage  goes  higher,  may  either  increase  or 
decrease  the  current  in  the  circuit  depending  upon  the 
power-factor  of  the  load  current  and  the  relative 
amount  of  the  leading  and  lagging  components  of  the 
current  in  the  circuit.  Any  change  in  the  current  of  the 
circuit  will  consequently  be  accompanied  by  a  corre- 
sponding change  in  the  I^R  loss.  In  fact,  these  sources 
of  additional  losses  may,  in  some  cases  of  long  circuits 
or  extensive  systems,  materially  contribute  toward  limit- 
ing the  transmission  voltage.    The  weight  ol  copper 


conductors,  from  which  their  cost  may  readily  be  cal- 
culated, is  given  in  Table  E-i.  As  an  insurance 
against  breakdown,  important  lines  frequently  are  built 
with  circuits  in  duplicate.  In  such  cases  the  cost  of 
conductors  for  two  circuits  should  not  be  overlooked. 

Table  E-1  contains  the  weights  of  bare  stranded 
copper  cables  per  1000  feet  of  circuit,  also  per  mile  of 
circuit.  For  the  purpose  of  facilitating  rapid  calcula- 
tion for  any  given  case,  the  weights  are  given  corre- 
sponding to  one,  two  and  three  conductors  for  these  two 
lengths  of  circuit. 

Reduced  Electric  Surges — The  better  insulation 
necessitated  by  higher  transmission  voltages  tends  to 
make  the  circuit  more  secure  against  ordinary  disturb- 
ances. Also  the  smaller  currents  resulting  with  the 
higher  voltages  cause  less  disturbance  in  the  circuit  in 
the  case  of  grounds,  short-circuits,  switchings,  light- 
ning and  other  disturbances. 

Less  Reactance  Volts  Drop — Since  the  current  cor- 
responding to  higher  transmission  voltages  goes  down 
as  the  voltage  goes  up,  the  voltage  necessary  to  over- 
come the  reactance  of  the  circuit  will  be  less,  and  the 
percentage  reactance  volts  much  less  for  higher  volt- 

TABLE  F— PRESENT  RELATIVE  COSTS  OF 

HIGH  TENSION  APPARATUS 

Expressed  in  Percent   (6600  Volt  Costs  Taken  as  100%) 


Trans- 
formers 

Switches 

Electro- 
lytic Ar- 
resters 

Insulators 


too 
100 


s> 


102 
100 


100  151 
100  135 


:> 


104 
100 


160 
185 


106 
100 


19s 
36s 


s> 


108 
100 


205 
430 


»> 


5> 


no 


320 
650 


125 
-15 


430 
1250 


150 

155 


640 
3S00 


175 

255 


1600 
5500 


20c 
420 


1900 
6500 


225 


2400 
7700 


ages.  Thus,  if  the  transmission  voltage  is  doubled,  the 
current  will  be  halved  and  for  the  same  spacing  of 
conductors  the  reactance  volts  drop  will  be  one  half,  re- 
sulting in  one  fourth  the  percentage  of  the  reactance- 
volts  drop. 

Cost  of  Transformers — If  the  transmission  voltage 
exceeds  13  200  volts,  banks  of  step-up  transformers  will, 
be  required  of  sufficient  capacity  to  transform  all  of 
the  kv-a.  to  be  transmitted.  A  still  greater  capacity  of 
step  down  transformers  will  be  lequired  to  reduce  the 
voltage  to  that  suitable  for  operating  motors  and  lights. 
In  some  cases  two  reductions  from  the  transmission 
circuit  voltage  may  be  required,  the  first  usually  re- 
ducing to  22  000,  1 1  000  or  6600  volts  for  general  dis- 
tribution and  the  second  reducing  from  the  general  dis- 
tribution voltage  to  the  proper  voltage  for  motors  and 
lights.  The  net  result  is  that  the  total  capacity  in  trans- 
formers connected  to  a  transmission  system  employing 
both  step  up  and  step  down  transformers  may  vary 
from  a  minimum  of  two  to  a  maximum  of  about  four 
times  the  kv-a.  transmitted  over  the  high-tension  cir- 
cuits. The  average  condition  we  will  assume  as  2.5 
times  the  kv-a.  to  be  transmitted. 

The  cost  of  power  transformers  at  the  present  time 


DETERMINATION  OF  FREQUENCY  AND  VOLTAGE 


47 


for  66  coo  volts  service  will  vary  between  $1.25  to  $3.00 
for  60  cycle  and  $2  to  $5  per  kv-a.  for  25  cycle  ser- 
vice, depending  upon  their  type  and  capacity.  The 
total  cost  per  kv-a.  of  transformers  on  a  system  would 
therefore  be  represented  by  approximately  2.5  times  the 
above  costs.  The  present  relative  costs  of  trans- 
formers for  different  voltages  are  given  in  Table  F. 
For  instance  if  the  transmission  voltage  is  increased 
from  33  000  to  66  000  volts  the  transformers  will  cost  in 
the  neighborhood  of  150  -^  115  or  31  percent  more 
than  they  would  cost  for  33  000  volts.  Knowing  the 
amount  of  power  to  be  transmitted,  an  approximate 
estimate  may  be  made  as  to  the  additional  cost  of  the 
necessary  transformers  for  a  higher  voltage. 

Cost  of  Insulators — Table  F  values  indicate  a  wide 
difference  in  the  cost  of  insulators  for  the  higher  volt- 


Efficiency — The  efficiency  of  transformers  will  be 
slightly  higher  for  the  lower  voltages. 

Small  Customers  —  The  furnishing  of  power  to 
small  customers  at  points  along  the  transmission  cir- 
cuits should  receive  careful  consideration.  The  cost  of 
switching  apparatus,  lightning  arresters  and  trans- 
formers required  to  permit  service  being  given  to  such 
customers  will  be  less  for  the  lower  voltage. 

Charging  Current  —  The  amount  of  current  re- 
quired to  charge  the  transmission  circuits  varies  ap- 
proximately as  the  transmission  voltage.  Therefore 
the  charging  current,  expressed  in  kv-a.  varies  ap- 
proximately as  the  square  of  the  voltage.  Thus  the 
charging  current  required  for  a  33  000  volt  circuit  is 
approximately  one  half  and  the  charging  kv-a.  one 
fourth  that  of  a  66  000  volt  circuit. 


TABLE  G— FORM  OF  TABULATION  FOR  DETERMINING  VOLTAGES 

AND  CONDUCTORS 

BASED  ON  THE  TRANSMISSION  OF  10  000  KV-A.  FOR  TEN  MILES  AT  80  PERCENT  POWER-FACTOR  LAQQINQ 

60  CYCLES,  THREE  PHASE 


VOLTAGE 

a: 

o< 

"-> 
co^ 

UJO 

< 

CONDUCTORS 

1 

VOLTAGE 
DROP  AT 

FIRST  COST 

ANNUAL 
OPERATING  COST 

FULL  LOAD 

CONDUCTORS 

AT  25  CTS.  PER 

POUND 

CO 

£< 

Si 

CO  0 

t- 

z 

Oco 

co^ 

l-H 
15 

Oco 

I 

S  UJ 
ZH 
H  CO 

-"< 

CO 

oc 

0 

CO 

z 

_l 
2 

0 

si 

UJ 

Z 

CO 

_l 

°i 

CO  < 

*^ 

00  0 
cc 
0 

is 

1- 

Ml 
0 

1% 

COI 

03O 

UJ 

TOTAL   |2rL0SS 

•INTEREST  ON 
FIRST  COST 
■  AT  6% 

DEPRECIATION 

ON  FIRST  COST 

AT  10% 

|2r  LOSSES 

AT  1  CT.  PER 

KW-HOUR 

1 

UJ 

0 

coir 

UJ  — 
(E 

UJ 

UJ 
CE 

a. 
0 

tr 

0 

> 

10.000 
KVA 

2500 
KVA 

TOTAL  LOSS 
PER  YEAR  IN 
KW-HOURS 

I2 

2 

lECO 
go: 

li.soo 

VSZi 

3  so 

SOO  000 

243  930 

i-n 

4  30 

.5.3 

27 

/  707470 

4.3 

2/.7 

17.S 

>60  9SZ 

*7SO0O 

*3  000 

'1  000 

*foo 

'nOiiZ 

'S4i3 

■'/4OS8'/7075fJ9</<l 

3  00  000 

Itk  L70 

i.tt> 

720 

9.0 

45 

2  867y50 

7.2 

22.7 

30 

3*  470 

7SO00 

3000 

1000 

900 

116  snd 

6994 

//6S7 

28  580 

47  331 

"000 

iioso 

3.S0 

/2S« 

16.1 

20 

5  /02  700 

/3.9 

24.2 

as 

20  5/2 

76  000 

3  000 

1000 

too 

100  4li 

6  02s 

10  041 

5/027^ 

67091 

Z7  fiOO 

n  702 

ZiZ 

iooooo 

lAi  670 

i.it 

403 

s.o 

2S 

1  S9S  700 

■^.0 

/2.8 

II 

36  670 

76  500 

3  000 

loSO 

1300 

/'«  420 

7  lOS 

//842 

/5  987 

3-»9* 

itooo 

82  030 

3.50 

720 

9.0 

-ts 

2  SS7  9SC 

7.2 

/3.i 

14 

XOS12 

76  SOO 

3  000 

loso 

1  200 

/02  362 

i,  136 

/0  226 

28  580 

44942 

•0 

.5/  t30 

.j.ii 

;/43 

14.3 

7/ 

+  i34  7M 

//.5 

14-1 

I7.S 

12  910 

76  500 

3  000 

1  050 

1200 

94  «0 

S  6B0 

9466 

45  348 

60  4f# 

33000 

I9  0S3 

ns 

»oo 

ii  100 

4.42 

-f06 

5.1 

25 

/  &09  6SC 

4.0 

6.S 

7.0 

16  275 

S3S00 

3  300 

/600 

1990 

/05  <« 

6  340 

10  565 

16  0?733002 

•2 

32  -4^0 

S.83 

8" 

/O.I 

60 

3  2/5  iSO 

8.0 

6.« 

/0.S 

8  117 

82500 

3  300 

1600 

/9S0 

974V 

5  850 

9749 

33  IS6  47  755 

>»+ 

20  430 

H.I 

IZfS 

I6.Z 

8/ 

5  HO  660 

/2.9 

7.1 

I4.S 

5  I07 

82  500 

3  300 

1  600 

1920 

94  4S7 

5  470 

944? 

5/407  66525 

44  000 

zs'io-t 

131 

»2 

32  460 

8.83 

A  St 

S.7 

29 

/  SOS  790 

4.6 

3.9 

6.0 

z  117 

90  000 

34i0 

2  300 

3960 

107721 

6  463 

/0  772 

IS  053 

3SZIS\ 

"■S 

/i  no 

/7.8 

9/6 

11.4 

S3 

3  639  780 

9.1 

■4.0 

9.S 

4  040 

90  000 

3450 

2  200 

3960 

10  3  tSO 

6319 

;0  3t5 

36   3»S529«| 

ages :  thus  the  increased  cost  of  66  000  volt  insulators 
above  the  cost  of  33  000  volt  insulators  is  stated  as 
3500  -f-  650  or  540  percent. 

Cost  of  Other  Apf>aratus — The  cost  of  lightning 
arresters,  high-tension  circuit  breakers  and  general  in- 
sulation increase  with  the  voltage.  -  The  increased  cost 
of  these  items,  however,  may  not  have  sufficient  weight 
to  materially  influence  the  selection  of  the  transmission 
voltage. 

Cost  of  Buildings  —  Lower  voltage  transformers, 
switching  equipment  and  lightning  arresters  require 
less  space  for  insulation.  If  this  apparatus  is  to  be 
placed  indoors,  the  cost  of  necessary  buildings  may  be 
less.  The  amount  of  real  estate  required  may  also  be 
less  in  case  of  the  lower  voltage. 

•  Relative  Cost  Values- — Table  F  contains  relative 
co.st  values  for  different  transmission  voltages.  They 
indicate  approximately  the  variation,  at  the  present 
time,  in  cost  of  the  principal  material  which  is  affected 
by  a  change  in  transmission  voltage.  Cost  values  are 
very  unstable  at  present  but  the  table  will  serve  in  a 
general  way  to  indicate  comparative  costs. 


Summary  —  In  deciding  upon  the  transmission 
voltage,  careful  and  full  consideration  should  be  given 
to  the  present  (or  probable  future)  voltage  of  any 
neighboring  or  adjacent  systems.  There  is  an  increas- 
ing tendency  to  combine  generating  and  transmission 
systems  for  purposes  of  economy,  and  insurance  against 
breakdown  in  service.  If  a  possible  future  consolida- 
tion is  not  kept  in  mind  when  selecting  the  transmission 
voltage,  a  voltage  may  be  decided  upon  which  would 
render  it  impossible  to  parallel  with  a  neiglAoring  sys- 
tem, except  through  connecting  transformers.  In  this 
case  the  transformers  of  the  two  systems  would  prob- 
ably not  be  interchangeal)le  for  service  on  either  system. 

If  the  contemplated  transmission  system  is  remote 
from  any  exi.sting  system,  a  study  of  the  initial  and  op- 
erating costs  should  be  made  corresponding  to  various 
sizes  of  conductors  and  to  various  assumed  transmission 
voltages.  A  suggested  tabulation  for  such  compari- 
sons is  shown  in  Table  G.  In  this  table,  it  is  assumed 
that  10  000  kv-a.  (8000  kw  at  80  percent  power-factor 
i:igging),  is  to  be  transmitted  a  distance  of  ten  miles 
at  60  cycles,  three-phase   for  ten  hours,   followed  by 


DETERMINATION  OF  FREQUENCY  AND  VOLTAGE 


2500  kv-a.  (2000  kw  at  80  percent  power-factor  lag- 
ging) for  14  hours.  Delta  spacing  is  assumed  of  three 
feet  for  the  lower  two  and  four  feet  for  the  higher  two 
voltages.  Raising  and  lowering  transformers  will  be 
required  of  an  assumed  total  capacity  of  2.5  X  10  000 
or  25  000  kv-a.  Conductors  of  hard  drawn  stranded 
copper  are  employed,  the  resistance  of  the  conductors 
being  taken  at  a  temperature  of  25  degrees  C.  from 
Table  II. 

The  cost  of  the  pole  or  tower  line,  the  righi  of 
way,  buildings  and  real  estate  for  buildings  is  not  in- 
cluded in  this  tabulation.  Neither  is  the  difference  in 
transformer  efficiencies  taken  into  account.  The  differ- 
ence in  these  items  will  not  be  sufficient  in  this  case 
greatly  to  influence  the  choice  of  the  transmission  volt- 
age, because  all  of  the  voltages  compared  are  relatively 
low.  Because  of  the  large  amount  of  power  to  be 
transmitted  a  comparatively  short  distance,  the  approxi- 
mate rule  of  1000  volts  per  mile  for  short  lines  does 
not  hold  true  for  this  problem. 

Assuming  for  the  sake  of  argument  that  the  price 
values  given  in  this  form  of  tabulation  are  approxi- 
mately correct  for  this  problem  and  that  there  are  no 
neighboring  transmission  systems,  then  the  problem  re- 
duces to  cost  economics. 

Since  both  the  first  and  operating  costs  in  Table  G 
are  higher  for  16  500  volts  than  they  are  for  22  000 
volts,  it  is  evident  that  16  500  volts  is  economically  too 
low  a  voltage. 

■  In  the  consideration  of  22  000  volts  it  will  be  seen 
that,  of  the  three  sizes  of  conductors,  the  largest  size 
(300000  circ.  mil.)  will  be  the  cheaper  in  the  end. 
Thus,  if  No.  000  were  selected,  the  first  cost  would  be 
$16159  less  than  for  30000  circ.  mil  conductors,  but 
the  operating  cost  (due  to  greater  loss  in  transmission) 
will  be  approximately  $10000  a  year  more.  For  a 
similar  reason  No.  o  conductors  will  be  disqualified. 

In  the  consideration  of  33  000  volts,  No.  00  con- 
ductors will  be  the  choice  and  in  the  consideration  of 
44000  volts.  No.  2  conductors  will  be  the  choice.  The 
choice  then  comes  down  to  the  following: — 


Voltage 
Transmission 

Conductors 

Total 

Cost 

First 

Annual 

Operating 

Cost 

22000 
33000 
44000 

300000  circ.  mils 
No.  00 
No.  2 

$118420 
105655 
107  727 

$34934 
33002 
35288 

It  will  thus  be  seen  that  a  voltage  of  33  000  volts 
and  No.  00  conductors  are  the  most  economical  of  those 
tabulated.  The  transmission  loss  will  be  5.1  percent, 
the  reactance  6.5  percent  and  the  voltage  drop  seven 
percent  at  full  load.    The  value  assigned  as  the  cost  per 


kw-hour  for  power  lost  in  transmission  will  obviously 
have  great  influence  in  determining  the  proper  economic 
size  of  conductors  for  any  given  transmission  voltage. 
The  cost  of  the  copper  will  have  a  relatively  greater  im- 
portance on  longer  lines.  As  a  matter  of  fact,  a  larger 
s'.ze  than  any  of  the  conductors  listed  in  Table  G  would 
be  still  more  economical,  under  the  conditions  given. 
There  have  been  numerous  mistakes  made  in  under-esti- 
mating the  ultimate  demand  for  electrical  power  and 
consequently  adopting  too  low  a  transmission  voltage. 
When  in  doubt  the  higher  voltage  will,  in  the  course  of 
time,  most  likely  justify  its  adoption  by  reason  of  fu- 
ture growth  not  apparent  at  the  time  the  choice  is  made. 
The  design  and  construction  of  transformers,  cir- 
cuit breakers,  lightning  arresters,  etc.  for  a  multiplicity 
of  high-tension  voltages  is  expensive.  The  manufac- 
turers of  such  apparatus  are  endeavoring  to  standardize 
transmission  voltages  for  the  purpose  of  minimizing  the 
number  of  designs  of  high-tension  apparatus.  This 
point  could  with  mutual  profit  be  taken  up  with  the 

TABLE  H-COMMON  TRANSMISSION  VOLTAGES 


Length  of  Line 

Voltages 

I  to      3  miles 

550  or       2200  volts 

3  to      5  miles 

2200  or       6600  volts 

5  to     10  miles 

6600  or     13200  volts 

10  to     15  miles 

13200  or    22000  volts 

15  to     20  miles 

22000  or    33000  volts 

20  to    30  miles 

33000  or    44000  volts 

30  to    50  miles 

44000  or    66000  volts 

SO  to    75  miles 

66000  or    88000  volts 

75  to  100  miles 

88000  or  1 10  000  volts 

100  to  150  miles 

no  000  or  132000  volts 

150  to  250  miles 

132000  or  154000  volts 

250  to  350  miles 

154000  or  220000  volts 

manufacturers  before  any  particular  voltage  is  decided 
upon. 

The  amount  and  cost  of  power  to  be  transmitted 
is  a  very  important  factor  in  determining  the  economic 
transmission  voltage.  For  average  conditions  isolated 
from  existing  transmission  lines  the  voltages  shown  in 
Table  H  have  been  quite  generally  used.  For  excep- 
tional cases,  exceptional  values  will  be  used.  For  ex- 
ample if  40000  kv-a.  is  to  be  transmitted  20  miles, 
66000  volts  or  higher  might  be  used.  On  the  other 
hand  if  a  very  small  amount  of  power  is  to  be  trans- 
mitted, lower  voltages  would  probably  be  selected. 

At  the  present  time  the  prospects  seem  bright  for 
the  standardization  of  the  following  "normal"  system 
voltages. 


44000 

132000 

66000 

154000 

88000 

♦187000 

110000 

220  000 

♦The  use  of  187000  volts  is  likely  to  occur  only  in  case  it 
i?  found  necessary  to  have  a  voltage  between  154  000  and  220  000 
volts. 


CHAPTER  VII 
PERFORMANCE  OF  SHORT  TRANSMISSION  LINES 

(EFFECT  OF  CAPACITANCE  NOT  TAKEN  INTO  ACCOUNT) 


THE  PROBLEMS  which  come  under  the  general 
heading  of  short  transmission  lines  are  those  in 
which  the  capacitance  of  the  circuit  is  so  small 
that  its  effect  upon  the  performance  of  the  circuit  may, 
for  all  practical  purposes,  be  ignored.  The  effect  of  ca- 
pacitance is  to  produce  a  current  in  leading  quadrature 
with  the  voltage,  usually  designated  as  charging  current. 
This  leading  component  of  current  in  the  conductor 
does  not  appear  in  the  load  current  at  the  receiving  end 
of  the  circuit.  It  is  zero  at  the  receiving  end  of  the 
circuit  but  increases  at  nearly  a  uniform  rate  as  the 
sending  end  of  the  circuit  is  approached,  at  which  point 
it  ordinarily  becomes  a  maximum. 

The  effect  of  this  charging  current  flowing  through 
the  inductance  of  the  circuit  is  to  increase  the  receiv- 
ing-end voltage  and  therefore  to  decrease  the  voltage 
drop  under  load.  Since  the  charging  current  is  2.4 
times  greater  for  a  frequency  of  60  cycles  than  it  is  for  a 
frequency  of  25  cycles,  its  effect  upon  the  voltage  regu- 
lation will  be  considerably  greater  at  60  cycles  than  at 
25  cycles.  The  effect  of  charging  current  upon  the 
voltage  regulation  will  also  increase  as  the  distance  of 
transmission  is  increased. 

If  the  circuit  were  without  capacitance,  there  would 
be  no  charging  current  and  consequently  the  mathe- 
matical and  the  two  graphical  solutions  (impedance 
methods)  which  follow  under  the  general  heading  of 
"short  transmission  lines"  would  all  produce  accurate 
results.  All  circuits,  however,  have  some  capacitance, 
and  as  the  length  or  the  frequency  of  the  circuit  in- 
creases, these  three  methods  will  therefore  yield  re- 
sults of  increasing  inaccuracy.  Some  engineers  con- 
sider these  impedance  methods  sufficiently  accurate  for 
circuits  20  to  30  miles  long  while  others  use  them  for 
still  longer  circuits.  To  act  as  a  guide.  Table  J  indicates 
the  error  in  the  supply  voltage  as  determined  by  these 
impedance  methods,  for  circuits  of  different  lengths 
corresponding  to  both  25  and  60  cycle  frequencies. 
These  three  impedance  methods  produce  practically  the 
same  results,  and  the  sending  end  voltage,  as  determined 
by  any  of  these  methods,  is  always  slightly  high.  In 
other  words  the  effect  of  the  charging  current  is  to 
reduce  the  voltage  necessary  at  the  sending  end,  for 
maintaining  a  certain  voltage  at  the  receiving  end  of 
the  circuit.  The  error  referred  to  below  for  the  three 
methods  is  expressed  in  percentage  of  the  receiving  end 
voltage.  Thus,  for  a  30  mile,  25  cycle  circuit,  the  error 
is  0.04  percent,  and  for  a  30  mile,  60  cycle  circuit  the 
error  is  0.2  percent.     If  an  error  of  0.5  percent  is  con- 


sidered permissible,  then  the  Dwight  or  the  Mershon 
Chart  methods,  or  the  corresponding  mathematical  so- 
lution, may  be  used  for  25  cycle  circuits  up  to  approxi- 
mately 125  miles,  and  for  60  cycles  circuits  up  to  ap- 
proximately 50  miles.  Of  course  these  impedance 
methods  may  be  used  for  still  longer  circuits  by  making 
proper  allowance  to  compensate  for  the  fundamental 
error. 

DIAGRAM   ILLUSTRATING  A  SHORT  TRANSMISSION   CIRCUIT 

Fig.  16  illustrates  the  relation  between  the  various 
elements  in  short  transmission  circuits,  when  the  effect 
of  capacitance  and  leakage  is  not  taken  into  account. 
The  current  flowing  in  such  a  circuit  meets  two  op- 
posing e.m.f's. ;  i.e.  of  resistance  in  phase  with  the  cur- 
rent and  reactance  in  lagging  quadrature  with  the  cur- 
rent. 

The  upper  part  of  Fig.  16  illustrates  such  a  circuit 
schematically  and  the  lower  part  vector ially.     The  volt- 

TABLE  J 


Length  of 
Circuit  (Miles) 

Error  in  Percentage  of 
Receiver  Voltage 

cycles 

60 
cycles  . 

20 

30 

50 

100 

200 

300 

-H>.02 

-f-0.04 

+ai 

-f04 

-fM 
+3-3 

-fa  10 
-fas 
-1-0.5 
+1.9 
-fao 
-fi8.o 

age  component  required  at  the  sending  end  to  overcome 
the  resistance  IR  of  the  circuit  is  indicated  in  the  vector 
diagram  by  a  short  line  parallel  with  the  base  line  /, 
representing  the  phase  of  the  current.  These  lines  are 
drawn  parallel,  since  the  resistance  voltage  drop  is  in 
phase  with  the  current.  The  voltage  component  re- 
quired at  the  sending  end  to  overcome  the  reactance  IX 
of  the  circuit  is  indicated  by  a  line  in  quadrature  or  at 
right  angles,  to  the  phase  of  the  current  The  reactance 
is  in  quadrature  with  the  current  for  the  reason  that  the 
rate  of  change  in  the  magnetic  field  (consequently  the 
e.m.f.  of  self-induction  or  reactance)  surrounding  the 
conductor  is  greatest  when  the  cufrent  is  passing 
through  zero.  The  hypotenuse  IZ  of  this  small  right 
angle  impedance  triangle  represents  the  impedance  volt- 
age of  the  circuit.  It  represents  the  direction  and  value 
of  the  resulting  voltage  necessary  to  overcome  the  com- 
bined effect  of  the  resistance  and  the  reactance  of  the 
circuit. 
The  relative  values  and  phases  of  the  receiving  and 


50 


PERFORMANCE  OF  SHORT  TRANSMISSION  LINES 


sending  end  voltages,  and  their  phase  relations  with  the 
current  /,  are  also  indicated  on  the  vector  diagram. 
This  diagram  is  plotted  for  a  receiving  end  load  based 
upon  80  percent  power- factor  lagging.  E,  represents  the 
value  of  the  voltage  required  at  the  sending  end  of  the 
circuit  to  maintain  the  voltage  £r  at  the  receiving  end, 
when  the  impedance  of  the  circuit  is  IZ  and  the  receiv- 
ing end  power-factor  is  80  percent  lagging.  The  phase 
angle  *,  indicates  the  amount  by  which  the  current  lags 
behind  the  voltage  at  the  sending  end;  cos  «,  being  the 
power-factor  of  the  load  as  measured  at  the  sending 
end.  Likewise  cos  «,  is  the  power-factor  of  the  load  at 
the  receiving  end. 

TAPS  TAKEN  OFF  CIRCUIT 

Usually  the  main  transmission  circuit  is  tapped  and 
power  taken  off  at  one  or  more  points  along  the  circuit. 
The  performance  of  such  a  circuit  must  be  calculated 
by  steps  thus: — Assume  a  circuit  200  miles  long  with 
10  000  kw  taken  off  at  the  middle  and  10  000  kw  at  the 
receiving  end.  From  the  conditions  known  or  assumed 
at  the  receiving  end,  calculate  the  corresponding  send- 


R         f         X 

Er 

NEUTRAL  _£_ 


P.  F.  OF 
LOAD=Co>  e. 


EQUIVALENT  TRANSMISSION  CIRCUIT  TO  NEUTRAL 


* ErCoie, ^^ 

VECTOR  DIAGRAM  OF  TRANSMISSION  CIRCUIT 

FIG.    16 — DIAGRAMS    FOR    SHORT    TRANSMISSION    LINES 

Impedance  method,  capacitance  effect  not  taken  into  account 

ing  end  conditions,  that  is  the  voltage,  power  and 
power-factor  at  the  substation  in  the  middle  of  the  cir- 
cuit. To  the  calculated  value  of  the  actual  power  in 
kilowatts  add  the  losses  at  the  substation  in  the  middle 
of  the  circuit.  Any  leading  or  lagging  component  in 
the  substation  load  current  must  also  be  added  alge- 
braically, in  order  to  determine  the  power-factor  at  the 
sending  side  of  the  substation.  This  will  then  be  the 
receiving  end  conditions  at  the  substation  in  the  middle 
of  the  circuit,  from  which  the  corresponding  conditions 
at  the  sending  end  of  the  circuit  may  be  calculated.  If 
the  sending  end  conditions  are  fixed,  and  the  receiving 
end  conditions  are  to  be  determined,  the  substation 
losses  will  in  such  case  be  subtracted  in  place  of  added. 

CABLE   AND   AERIAL   LINES    IN    SERIES COMPOSITE    LINES 

In  some  cases  it  is  necessary  to  place  part  of  a 
transmission  circuit  underground,  and  in  other  cases  it 
may  be  desirable  to  use  two  or  more  sizes  of  conductors 
in  series.  The  result  will  be  that  the  circuit  constants 
will  be  different  for  the  various  sections.  If  the  effect 
of  capacitance  be  neglected,  the  combined  circuit  may 


be  treated  as  a  single  circuit  having  a  certain  total  re- 
sistance R  and  a  total  reactance  X. 

PROBLEMS 

Later  a  table  will  be  presented  listing  a  large  num- 
ber of  transmission  circuits  from  20  to  500  miles  long, 
at  both  25  and  60  cycles  operating  at  from  10  000  to 
200  000  volts.  These  problems  are  numbered  from  i  to 
64.  When  a  reference  is  made  in  the  following  to  some 
problem  number  it  will  refer  to  one  of  this  list  of  prob- 
lems. 

SYMBOLS 

The  symbols  which  will  be  employed  in  the  follow- 
ing treatment  are  given  below: — 

FOR  LOAD  CONDITIONS 

Kv-Or    =  (total)  at  receiving  end. 
Kv-ciro  =  (one  conductor  to  neutral)  at  receiving  end. 
Kv-a,    =  (total)  at  sending  end. 
Kv-OsB  =  (one  conductor  to  neutral)  at  sending  end. 
Kwr    =  Kw  (total)  at  receiving  end. 
Kwra  =  Kw  (one  conductor  to  neutral)  at  receiving  end. 
Kw,    =  Kw  (total)  at  sending  end. 
Kw,n  =  Kw  (one  conductor  to  neutral)  at  sending  end. 
E,    =  Voltage  between  conductors  at  receiving  end. 
£rii   =  Voltage  from  conductors  to  neutral  at  receiving 

end. 
E,    :=  Voltage  between  conductors  at  sending  end. 
£»«   =  Voltage  from  conductors  to  neutral  at  sending 
end. 
I,    =  Current  in  amperes  per  conductor  at  receiving 

end. 
/.    =  Current  in  amperes  per  conductor  at  sending  end. 
Cos  6,    =  Power-factor  at  receiving  end. 
Cos  6,    =  Power-factor  at  sending  end. 

FOR  ZERO   LOAD  CONDITIONS 

The  symbols  corresponding  to  zero  load  conditions 
are  as  indicated  above  for  load  conditions  with  the  ad- 
dition of  a  sub  zero. 

THE   FUNDAMENTAL  OR  LINEAR  CONSTANTS 

The  fundamental,  or  "linear  constants"  of  the  cir- 
cuit for  each  conductor  per  unit  length  are  represented 
as  follows: — 

r  :=  Linear  resistance  in  ohms  per  conductor  mile  (taken 

from  Table  II) 
X  =  Linear  reactance  in  ohms  per  conductor  mile   (taken 

from  Table  IV  or  V) 
b  =  Linear  capacitance  susceptance  to  neutral  in  mhos  per 

conductor  mile  (taken  from  Table  IX  or  X> 
g  =  Linear  leakage  conductance  to  neutral  in  mhos  per 
conductor  mile.  (This  represents  the  direct  escape 
of  active  power  through  the  air  between  conductors 
and  of  active  power  leakage  over  the  insulators. 
These  losses  must  be  estimated  for  conditions 
similar  to  these  of  the  circuit  under  consideration. 
For  all  lines  except  those  of  great  length  and  high 
voltage  it  is  common  practice  to  disregard  the 
effects  of  leakage  or  corona  loss  and  to  take  g  as 

equal  to  zero.  

z  ^  Linear  impedance  =   ]/  r*  -|-  x* 
y  =  Linear  admittance  =  \/  g'  -j-  b' 
If  the  length  of  each  conductor  of  the  circuit  in 
unit  length  is  designated  as  /  we  have 

rl  =  Total  resistance  in  ohms  per  conductor  =  R 
xl  :=  Total  reactance  in  ohms  per  conductor  =  X 
hi  =  Total   susceptance   in   mhos  per  conductor  to  neutral 

D 

gl  =  Total  conductance  in  mhos  per  conductor  to  neutral 
=  G 


PERFORMANCE  OF  SHORT  TRANSMISSION  LINES 


51 


then,  

Z  =  y  R'  +  X'  ohms 
and,  Y  =  \/G'  +  B'  mhos 

IR  =  Voltage  necessary  to  overcome  the  resistance. 
IX  =  Voltage  necessary  to  overcome  the  reactance. 
IZ  ^  Voltage  necessary  to  overcome  the  impedance. 

METHODS    FOR   DETERMINING   THE    CONSTANTS   OF   THE 
CIRCUIT 

Several  different  methods  for  determining  the  fim- 
damental  constants  of  the  circuit  are  in  use.  These 
methods  are  illustrated  below. 

Problem — Find  the  resistance  volts  IR  and  the  re- 
actance volts  IX  in  percent  of  delivered  volts  £,  for  the 
following  conditions: — lOO  kw  active  power  to  be  de- 
livered at  looo  volts,  three-phase,  60  cycles,  over  three 
No.  0000  stranded,  hard  drawn,  copper  conductors,  cir- 
cuit one  mile  long,  with  a  symmetrical  delta  arrange- 
ment of  conductors,  two  foot  spacing,  the  temperature 
being  taken  as  25  degrees  C. 

Resistance   of   one   mile  of   single  conductor  =  0.277 

ohm  (from  Table  II) 

Reactance   of   one   mile   of   single  conductor  =  0-595 

ohm  (from  Table  V) 

Method  No.  i — When  three-phase  circuits  first 
came  into  use,  it  was  customary  (and  correct),  in  de- 
termining the  loss  and  voltage  regulation,  to  consider 
them  equivalent  to  two  single-phase  circuits,  each  single- 
phase  circuit  transmitting  one-half  the  power  of  the 
three-phase  system.  This  practice  is  still  followed  by 
some  engineers;  thus: — 

50000 


1000 


50  amp.  per  conductor  for  each  single-phase  cir- 
cuit. 

0.277  X  2  X  SO  ^  .  ■       ,  J- 
'-'-^ — £i-^_  ^   jQQ  _  2.77  %    resistance  volts  drop  of 

'"^^^  single-phase  circuit. 

0-595  X  2  X  50  ^   jQQ  _  5  55%    reactance  volts  drop  of 

'°'-'°  single-phase  circuit. 

Method  No.  z  consists  of  treating  the  case  as  a 
straight  three-phase  problem.     Thus: 

—, •    =    ';7-7'?   amperes   per   conductor   of   three- 

1000  X  1-732  phase  circuit. 

0.277  X  1-732  X  57-73    y^    ^^   ^    ^.^.^^^    resistance  volts 
i°oo  drop    of    three- 

phase  circuit. 
0-595  X  1-732  X  57-73...  y    ^^   _  %    reactance   volts 

1000  ^  ='""       drop    of    three- 

phase  circuit. 

Method  No.  3  consists  in  assuming  one-third  the 
total  power  transmitted  over  one  conductor  with  neutral 
or  ground  return  (resistance  and  reactance  of  return 
being  taken  as  zero).  Such  an  equivalent  circuit  is 
shown  by  diagram  in  the  upper  part  of  Fig.  16.  Thus 
the  circuit  constants  for  the  above  problem  would  be 
determined  as  follows : — 

100  000 
Watts  per  phase  =   — - —  —  33  333  waits. 

Volts  to  neutral  =  1000  X  O-S774  or  S77A  volts. 

33333.   _   ^^^^   amperes   per    cotiducior;    {smme   as   for 

577-4  method  No.  2) 

0.277  X  57-74.  ^   100  —  2.77%    resistance  volts  drop  of 

577-4  three-phase  circuit. 

0-595  X  57-74    y    ,00  _  5.95%    reactance  volts  drop   of 
577-4  three-phase  circuit. 


It  will  be  seen  that  all  three  methods  produce  the 
same  results.  Method  No.  j  seems  the  most  readily 
adaptable  to  various  kinds  of  transmission  systems  and 
will  be  used  exclusively  in  the  treatment  of  the  problems 
which  will  follow. 

APPLICATION  OF  THE  TABLES 

Numerous  tables  of  constants,  charts,  etc.,  have 
been  presented,  and  a  few  more  will  follow.  Chart  II 
plainly  indicates  the  application  of  these  tables,  etc.  to 
the  calculation  of  transmission  circuits  and  the  sequence 
in  which  they  should  be  consulted. 

GRAPHICAL  vs.    MATHEMATICAL  SOLUTIONS 

At  the  time  of  the  design  of  a  transmission  circtiit 
the  actual  maximum  load  or  power- factor  of  the  load 
that  the  circuit  will  be  called  upon  to  transmit  is  sel- 
dom known.  An  unforseen  development  leading  to  an 
increased  demand  for  electrical  energy  may  result  in  a 
greatly  increased  load  to  be  transmitted.  The  acttial 
length  of  a  circuit  (especially  when  located  in  a  hilly 
or  rolling  country)  is  never  known  with  mathematical 
accuracy.  Moreover,  the  actual  resistance  of  the  con- 
ductors varies  to  a  large  extent  with  temperature  varia- 
tions along  the  circuit. 

When  it  is  considered  that  there  are  so  many  in- 
determinate variables  which  vitally  affect  the  per- 
formance of  a  transmission  circuit,  it  would  seem  that 
a  comparatively  long  and  highly  mathematical  solution 
for  determining  the  exact  performance,  necessarily 
based  upon  rigid  assumptions,  is  hardly  justified.  In 
many  cases  the  economic  loss  in  transmission  will  de- 
termine the  size  of  conductors  and,  if  the  circuit  is  very 
long,  synchronous  machinery  is  likely  to  be  employed 
for  controlling  the  voltage. 

Mathematical  solutions  have  one  very  important 
virtue,  in  that  they  provide  an  entirely  different  but 
parallel  route  in  the  solution  of  such  problems,  and 
therefore  are  valuable  as  a  check  against  serious  errors 
in  the  results  obtained  by  the  more  simple  graphical  so- 
lutions. 

In  the  following  treatment,  simple  but  highly  ac- 
curate graphical  solutions  will  be  first  presented,  for 
determining  the  performance  not  only  of  short  trans- 
mission lines,  but  also  for  long  lines.  For  short  lines 
the  Dwight  and  the  Mershon  charts  will  be  used.  For 
long  lines,  where  the  effect  of  capacitance  must  be  ac- 
curately accounted  for,  the  Wilkinson  Charts,  supple- 
mented with  vector  diagrams  will  be  used.  These  three 
forms  of  graphical  solutions  will,  when  correctly  ap- 
plied to  any  power  transmission  problem,  produce  re- 
sults in  which  the  error  will  be  much  less  than  that  due 
to  irregularities  in  line  construction  and  inaccurate  as- 
sumptions of  circuit  constants.  These  three  graphical 
solutions  will  in  each  case  be  followed  by  mathematical 
solutions.  In  the  case  of  short  lines  the  usual  formulas 
employing  trigonometric  functions  will  be  employed, 
and  in  the  case  of  long  lines  the  convergent  series,  and 
two  different  forms  of  hyperbolic  solutions  will  be  em- 
ployed. 


52 


PERFORMANCE  OF  SHORT  TRANSMISSION  LINES 


GRAPHICAL  SOLUTION 

When  the  receiving  end  load  conditions,  that  is, 
the  voltage,  the  load  and  the  power-factor  are  known, 
the  IR  volts  required  to  overcome  the  resistance  and 
the  IX  volts  required  to  overcome  the  reactance  of  the 
circuit,  may  be  readily  calculated. 

On  a  piece  of  plain  paper  or  cross-section  paper 
divided  into  tenths,  a  vector  diagram  of  the  current 
and  of  the  various  voltage  drops  of  the  circuit  may  be 
laid  out  to  a  convenient  scale.  Whichever  kind  of 
paper  is  used,  the  procedure  will  be  as  in  the  following 
example. 

Single-Phase  Problem — Find  the  voltage  at  the 
sending  end  of  a  single-phase  circuit  i6  miles  long,  con- 
sisting of  two  stranded,  hard  drawn  No.  ocxx>  copper 
conductors  spaced  three  feet  apart.  Temperatures  taken 
as  25  degrees  C.  Load  conditions  at  receiving  end 
assumed  as  4000  kv-a  (3200  kw  at  80  percent  power- 
factor  lagging)  20  000  volts,  single-phase,  60  cycles. 
4000 


Kv-ar„  =- 


Era 


2 
20  000 


2000  kv-a  to  neutral. 


10  000  volts  to  neutral. 


2  000  000 
Ir    = —  —  ^  200  amperes  per  conductor. 

The  fundamental  constants  per  conductor  are: — 
i?  =  16  X  0.277  'from  Table  II)  =  4.432  ohms 
X  =  16  X  0.644  (from  Table  V)  ^  10.304  ohms 
and  IR  =  200  X  4.432  ^  886  volts  resistance  drop 
886 
=  ,„,,^    X  i«o  =  8.86  percent 
10000  "^ 

IX  =  200  X  10.304  =  2061  volts  reactance  drop 
2061     ^ , 

=  ,„„^    X  100  —  20.61  percent 
loooo  ' 

Having  determined  the  above  values  a  vector  dia- 
gram may  be  made  as  follows: — 

Draw  an  arc  quadrant  having  a  radius  of  10  000 
(the  receiving  end  voltage  to  neutral)  to  some  con- 
venient scale,  as  shown  in  Fig.  17.  The  radius  which 
represents  the  base,  or  horizontal  line  will  be  assumed 
as  representing  the  phase  of  the  current  at  the  receiv- 
ing end  of  the  circuit.  Divide  this  base  line  into  ten 
equal  parts.  These  ten  divisions  will  then  correspond 
to  loads  of  corresponding  power-factors.  Since  a  load 
has  been  assumed  having  a  power-factor  of  80  percent 
lagging,  draw  a  vertical  line  from  the  0.8  division  on 
the  base  line,  until  it  intersects  the  arc  of  the  circle. 
From  this  point  of  intersection  draw  a  line  to  the  right 
and  parallel  with  the  base  line.  To  the  same  scale  as 
that  plotted  for  the  receiver  voltage  (10  000)  measure 
off  to  the  right  886  volts  to  D.  This  is  the  voltage 
which,  as  determined  above  is  required  to  overcome  the 
resistance  of  one  conductor  of  the  circuit.  It  is  some- 
times stated  as  the  voltage  consumed  by  the  line  re- 
sistance. It  will  be  noted  that  this  voltage  drop  is  in 
phase  with  the  current  at  the  receiving  end.  From  this 
point  lay  off  vertically,  and  to  the  same  scale,  2061  volts 
which  is,  as  determined  above,  the  volts  necessary  to 
overcome  the  reactance  of  one  conductor  of  the  circuit. 
This  is  sometimes  stated  as  the  voltage  consumed  by  the 
line  reactance.     Connect  this  last  point  by  a  straight 


CHART  II.— APPLICATION  OF  TABLES  TO 
SHORT  TRANSMISSION  LINES 

(EFFECT  OF  CAPACITANCE  NOT  TAKEN  INTO  AC- 
COUNT)     OVER  HEAD  BARE  CONDUCTORS 


Starting  with  the  kv-a.,  voltage  and  power-factor  at 
the  receiving  end  known. 


QUICK  ESTIMATING  TABLES  XII  TO  XXI  INC. 

From  the  quick  estimating  table  corresponding  to  the 
voltage  to  be  delivered,  determine  the  size  of  the  con- 
ductors corresponding  to  the  permissible  transmission  loss. 


HEATING  LIMITATION— TABLE  XXIII 

If  the  distance  of  transmission  is  short  and  the  amount 
of  power  transmitted  very  large  there  is  a  possibility  of 
overheating  the  conductors — to  guard  against  such  over- 
heating the  carrying  capacity  of  the  conductors  contem- 
plated should  be  checked  by  this  table. 


CORONA  LIMITATION— TABLE  XXII 

If  the  transmission  is  at  30000  volts,  or  higher,  this 
table  should  be  consulted  to  avoid  the  employment  of  con- 
ductors having  diameters  so  small  as  to  result  in  excessive 
corona  loss. 


RESISTANCE— TABLES  I  AND  II 
From  one  of  these  tables  obtain  the  resistance  per 
unit  length  of  single  conductor  corresponding  to  the  maxi- 
mum operating  temperature — calculate  the  total  resistance 
for  one  conductor  of  the  circuit — if  the  conductor  is  large 
(250000  circ.  mils  or  more)  the  increase  in  resistance  due 
to  skin  effect  should  be  added. 


PR  TRANSMISSION  LOSS 

Calculate  the  PR  loss  of  one  conductor  by  multiplying 
its  total  resistance  by  the  square  of  the  current — to  obtain 
the  total  loss  multiply  this  result  by  the  number  of  con- 
ductors of  the  circuit. 


REACTANCE— TABLES  IV  AND  V 

From  one  of  these  tables  obtain  the  reactance  per  unit 
length  of  single  conductor.  Calculate  the  total  reactance 
for  one  conductor  of  the  circuit.  If  the  reactance  is  ex- 
cessive (20  to  30  percent  reactance  volts  will  in  many  cases 
be  considered  excessive)  consult  Table  VI  or  VII.  Hav- 
ing decided  upon  the  maximum  permissible  reactance  the 
corresponding  resistance  may  be  found  by  dividing  this 
reactance  by  the  ratio  value  in  Table  VI  or  VII.  When 
the  reactance  is  excessive,  it  may  be  reduced  by  instaljing 
two  or  more  circuits  and  connecting  them  in  parallel,  or 
by  the  employment  of  three  conductor  cables.  Using 
larger  conductors  will  not  materially  reduce  the  reactance. 
The  substitution  of  a  higher  transmission  voltage,  with  its 
correspondingly  less  current,  will  also  result  in  less  react- 
ance. 


GRAPHICAL  SOLUTION 
A  simple  graphical  solution,  as  described  in  the  text, 
may  be  made  by  which  the  kv-a,  the  voltage  and  the  power- 
factor  at  the  sending  end  of  the  circuit  may  be  determined 
graphically.  Or  the  voltage  at  the  sending  end  may  be 
determined  graphically  by  the  use  of  either  the  Dwight  or 
the  Mershon  chart.  With  the  Mershon  chart  the  power- 
factor  at  the  sending  end  may  be  read  directly  from  the 
chart. 


MATHEMATICAL  SOLUTION 
As   a   precaution   against   errors   the   results  obtained 
graphically  should  be  checked  by  a  mathematical  solution, 
in  cases  where  accuracy  is  essential. 


PERFORMANCE  OF  SHORT  TRANSMISSION  LINES 


53 


line  with  the  center  E  of  the  arc.  The  length  of  this 
line  ES  represents  the  voltage  to  neutral  at  the  sending 
end  which,  for  this  problem,  is  1 1  998  volts.  The  dis- 
tance this  line  extends  beyond  the  arc  represents  the 
drop  in  voltage  for  one  conductor  of  the  circuit.    The 


voltage  drop  for  this  problem  is 


1998 


X  100  =  19.98 


10000 
percent  of  the  receiving  end  voltage. 

The  phase  difference  between  the  current  and  the 
voltage  at  the  receiver  end  is  e^  =  36°  52'.  This  is  the 
angle  whose  cosine  is  0.8  corresponding  to  a  power-fac- 
tor at  the  receiving  end  of  80  percent.  Likewise  the 
phase  difference  between  the  receiving  end  current  and 
the  sending  end  voltage  is  *,  ^  42°  13'  corresponding 
to  a  power-factor  at  the  supply  end  of  74.06  percent. 
The  difference  in  these  two  phase  angles  (5°  21')  repre- 
sents the  difference  in  the  phase  of  the  voltages  at  the 
sending  and  receiving  ends  of  the  circuit.     The  power- 

I 


I9»a  VOLTS  CONDUCTOR  DROP 

e206l  VOLTS  REACTANCE  DROP 
2S44  VOLTS  IMPEDANCE  DROP 


eee  volts  resistance  drop 

POSITION   FOR    load 
OF  80%  UAQGING. 


/i 


/* 

/.?" 


$ 


ey 


0.a         0.4         06         0.6         0.7         0.8 
POWER-FACTOR     OF   LOAD 


>  CURRENT 


FIG.    17 — GRAPHICAL    SOLUTION    FOR    A    SHORT   TRANSMISSION    LINK 

Capacitance  effect  not  taken  into  account. 

factor  at  the  sending  end  of  the  circuit  may  be  readily 
obtained  by  dropping  a  vertical  line  down  from  the  point 
where  the  line  representing  the  sending  end  voltage  £5" 
intersects  the  arc  of  the  circle,  to  the  base  line 
representing  the  phase  of  the  receiving  end  current. 
Such  a  line  will  correspond  to  a  power-factor  of  74.06 
percent.  This  assumption  that  the  vector  representing 
the  direction  of  the  receiving  end  current  also  repre- 
sents the  direction  of  the  sending  end  current  is  upon 
the  basis  that  the  circuit  is  without  capacitance.  It, 
therefore,  is  permissible  only  with  short  lines. 

In  Fig.  17  the  location  of  the  impedance  triangle 
is  also  indicated  (by  broken  lines)  in  positions  corre- 
sponding to  a  receiving  end  load  of  100  percent  power- 
factor  ;  and  also  for  a  receiving  end  load  of  zero  lagging 
power- factor.  It  is  interesting  to  note  that  in  the  case 
of  100  percent  power-factor  the  resistance  drop  (at 
right  angle  to  the  arc)  has  a  maximum  effect  upon  the 
voltage  drop;  whereas  the  reactance  drop  (nearly  paral- 
lel with  the  arc)  has  a  minimum  effect  upon  the  volt- 


age drop.  At  zero  lagging  power-factor  load  just  the 
reverse  is  true;  namely  the  resistance  drop  is  nearly 
parallel  with  the  arc  and  causes  a  minimum  voltage 
drop,  while  the  reactance  is  at  right  angles  and  produces 
a  maximum  effect  upon  the  voltage  drop. 

VOLTAGE   AT   SENDING   END   AND   LOAD  AT  RECEIVING 
END  FIXED 

In  cases  of  feeders  to  be  tapped  into  main  trans- 
mission circuits,  the  voltage  at  the  sending  end  is  usually 
fi.xed.  It  may  be  desired  to  determine  what  the  volt- 
age will  be  at  the  receiving  end  corresponding  to  a  given 
load.     This  may  be  obtained  graphically  as  follows: — 

Draw  a  horizontal  line  which  will  be  assumed  to 
represent  the  phase  of  the  current.  (Fig.  17)  Since 
the  power-factor  of  the  load  at  the  receiving  end  is 
known,  the  angle  whose  cosine  corresponds  may  be  ob- 
tained from  Table  K.  This  angle  represents  the 
phase  relation  between  the  current  and  the  voltage  af 
the  receiving  end  of  the  circuit.  For  the  problem  illus- 
trated by  Fig.  17  this  angle  is  36°  52',  corresponding  to 
a  power-factor  of  80  percent.  Having  determined  this 
angle,  draw  a  second  radial  line  intersecting  the  current 
vector  at  the  angle  corresponding  to  the  receiving  end 
load  power-factor.  This  second  line  will  then  repre- 
sent the  direction  of  the  voltage  at  the  receiving  end 
of  the  circuit.  If  the  load  power- factor  is  lagging,  this 
line  will  be  in  the  forward  direction,  and  if  the  load 
power-factor  is  leading  it  will  be  in  the  backward  di- 
rection from  the  current  vector.  Now  with  the  inter- 
section of  the  current  and  voltage  vectors  as  a  center, 
draw  an  arc  of  a  circle  to  some  suitable  scale,  repre- 
senting the  voltage  at  the  sending  end.  Calculate  the 
voltage  necessary  to  overcome  the  resistance,  and  also 
that  necessary  to  overcome  the  reactance  of  the  circuit 

Draw  a  right  angle  impedance  triangle  to  the  same 
scale,  using  the  resistance  volts  as  a  base.  Cut  out  the 
impedance  triangle  to  its  exact  size.  Keeping  the  base 
of  the  triangle  (resistance  voltage)  in  a  horizontal  posi- 
tion (parallel  with  the  current  vector)  move  the  triangle 
over  the  diagram  in  such  a  manner  that  its  apex  follows 
the  arc  of  the  circle  representing  the  numerical  value 
of  the  voltage  at  the  sending  end.  Move  the  triangle 
up  or  down  until  a  position  is  found  where  it  makes 
connection  with  the  vector  representing  the  voltage  at 
the  receiving  end.  This  is  then  the  correct  position  for 
the  impedance  triangle,  and  the  receiving  end  voltage 
may  be  scaled  off. 

GRAPHICAL  SOLUTION   BY  THE  MERSHON  CHART 

The  above  graphical  solution  is  that  employed  in 
the  well  known  chart  which  Mr.  Ralph  D.  Mershon 
early  presented  to  the  electrical  profession,  and  which 
is  reproduced  as  Chart  III.  The  Mershon  Chart  is 
simply  a  diagram  on  cross-section  paper  with  vertical 
and  horizontal  subdivisions  each  representing  one  per- 
cent of  receiving  end  voltage.  On  this  chart  a  number 
of  concentric  arcs  are  drawn,  representing  voltage  drops 
up  to  40  percent.  After  the  reactance  and  the  resist- 
ance volts  have  been  calculated  and  expressed  in  per- 


54 


PERFORMANCE  OF  SHORT  TRANSMISSION  LINES 


CHART   lll-MERSHON   CHART 


DIRECTIONS    FOR    USING    CHART 

BY  MEANS  OF  THE  TABLES  CALCULATE  THE  RESISTANCE  VOLTS  AND  THE  REACTANCE. 
^VOLTS  IN  THE  LINE,  AND  FIND  WHAT  PER  CENT  EACH   IS  OF  THE  E.M.F.  DELIVERED 
■^AT  THE  END  OF  THE  LINE.  STARTING  FROM  THE   POINT  ON  THE  CHART  WHERE 
^THE  VERTICAL  LINE  CORRESPONDING  WITH  POWER  FACTOR  OF    THE    LOAD 
JNTERSECTS  THE  SMALLEST  CIRCLE.  LAY  OFF  IN   PER  CENT  THE  RESIS- 
iJ-ANCE   E.M.F.  HORIZONTALLY  AND  TO  THE  RIGHT^  FROM   THE  POINT 
yTHUS  OBTAINED  LAY  OFF  UPWARD  IN   PER  CENT  THE  REACTANCE- 
THE  CIRCLE  ON  WHICH  THE  LAST  POINT  FALLS  GIVES 
DROP  IN   PER  CENT  OF  THE  E.M.F.  DELIVERED  AT  THE 
THE  LINE. 

EXAMPLE 

VTHUS  A  CIRCUIT  HAVING  5  PER  CENT  RESISTANCE 

D  to  PER  CENT  REACTANCE  VOLTS   HAS  WITH 

LOAD  OF  lOO    PER  CENT   POWER    FACTOR   5^ 

PER  CENT  VOLTAGE  DROP  OR  WITH  A  LOAD 

OF  ao  PER  CENT  POWER   FACTOR   10  PER 

CENT  VOLTAGE  DROP. 

RESISTANCE  DROP_ 


.6  .7  .8 

LOAD  POWER  FACTORS 


0  10  20  30 

DROP  IN  PER  CENT  OF  E.M.F.  DELIVERED 


PERFORMANCE  OF  SHORT  TRANSMISSION  LINES 


S5 


cent  of  Ef  the  impedance  triangle  is  traced  upon  the 
chart  and  the  voltage  drop  in  percentage  of  £,  is  read 
directly  as  indicated  by  the  directions.  All  values  on  the 
chart  are  expressed  in  percent  of  the  receiving  end  volt- 
age. 

Single-Phase  Problem — Taking  the  resistance  volt- 
age as  8.86  percent  and  the  reactance  voltages  20.61  per- 
cent of  the  receiving  end  voltage,  for  the  above  single- 
phase  problem,  (Fig.  17)  and  tracing  these  values  upon 
the  Mershon  Chart  for  a  receiving  end  load  of  80  per- 
cent power-factor  lagging,  the  voltage  drop  is  de- 
termined as  19.9  percent.  The  calculated  value  being 
19.98  percent,  the  error  by  the  chart  is  seen  to  be 
negligible. 

WHEN   THE  SENDING  END   CONDITIONS  ARE  FIXED 

When  the  conditions  at  the  sending  end  are  fixed 
and  those  at  the  receiving  end  are  to  be  determined, 
the  solving  of  the  problem  by  the  Mershon  Chart  is 
more  complicated.  In  such  cases,  it  is  usual  to  estimate 
what  the  probable  receiving  end  condition  will  be. 
From  these  estimated  receiving  end  conditions,  deter- 
mine by  the  chart  the  corresponding  sending  end  condi- 
tions. If  the  conditions  as  determined  by  this  assump- 
tion are  materially  'different  from  the  known  conditions, 
another  assumption  should  be  made.  The  correspond- 
ing sending  end  conditions  should  then  be  checked  with 
the  known  conditions.  Several  such  trials  will  usually 
be  necessary  to  solve  such  problems. 

GRAPHICAL    SOLUTION    BY    THE    DWIGHT    CHART 

Mr.  H.  B.  D wight  has  worked  up  a  straight  line 
chart,  shown  as  Chart  IV,  in  which  the  resistance  and 
the  reactance  of  the  circuit  have  been  taken  into  ac- 
count through  the  medium  of  spacing  lines  marked  for 
various  sizes  of  conductors.*  The  use  of  this  chart 
does  not,  therefore,  require  the  calculation  of  the  re- 
sistance and  reactance  or  the  use  of  tables  of  such  con- 
stants. The  Dwight  Chart  is  also  constructed  so  as  to 
be  applicable  to  loads  of  leading  as  well  as  to  loads  of 
lagging  power-factors,  whereas  the  Mershon  chart,  as 
generally  constructed,  is  applicable  to  loads  of  lagging 
power-factor  only.  However  the  Mershon  Chart  can 
be  made  applicable  for  the  solving  of  problems  of  lead- 
ing as  well  as  lagging  power-factor  loads  by  extending 
it  through  the  lower  right-hand  quadrant.  The  appli- 
cation of  synchronous  condensers  frequently  gives  rise 
to  loads  of  leading  power-factor.  The  Dwight  Chart 
is  well  adapted  to  the  solution  of  such  circuits.  Still 
another  feature  of  this  chart  is  that  formulas  are  given 
which  take  capacitance  effect  into  account  with  suffi- 
cient accuracy  for  circuits  with  a  length  up  to  approxi- 
mately 100  miles. 

Single-Phase  Problem — Find  the  voltage  at  the 
sending  end  of  a  single-phase  circuit  16  miles  long,  con- 
sisting of  two  stranded,  hard-drawn,  No.  0000  copper 
conductors,    spaced    three    feet    apart.      Temperature 


*The  basis  of  the  construction  of  this  chart  13  described  in 
the  Journal  for  July,  1915,  p.  306.  * 


taken  as  25  degrees  C.  Load  condition  at  receiving  end 
assumed  as  4000  kv-a  (3200  kw  at  80  percent  power- 
factor  lagging)  20  000  volts  single-phase,  60  cycles. 

From  Table  II  the  resistance  of  No.  0000  stranded, 
hard-drawn,  copper  conductors  at  25  degrees  C.  is 
fotmd  to  be  0.277  ohm  per  wire  per  mile.  Lay  a 
straight  edge  across  the  Dwight  Chart  from  the  resist- 
ance value  per  mile  0.277  (^s  read  on  the  lower  half 
of  the  vertical  line  to  the  extreme  right)  to  the  spacing 
of  three  feet  for  copper  conductors  and  60  cycles  at  the 
extreme  left.  Along  this  straight  edge  read  factor 
V  =  0.62,  corresponding  to  a  lagging  power-factor  of 
80  percent.  This  factor  V  is  equivalent  to  the  change 
in  receiving  end  voltage  per  total  ampere  per  mile  of 
circuit,  due  to  the  line  impedance. 

It  will  be  noted  that  opposite  the  resistance  values 
(extreme  right  vertical  line)  is  placed  the  correspond- 
ing sizes  of  copper  and  aluminum  conductors  on  the 
basis  of  a  temperature  of  20  degrees  C.  If  the  tempera- 
ture is  assumed  to  be  20  degrees  C.  it  will  not  be  neces- 
sary to  consult  a  table  of  resistance  values.  In  such  a 
case,  the  straight  edge  would  simply  be  placed  over  the 
division  of  the  vertical  resistance  line  corresponding  to 
the  size  and  material  of  conductors.  Marking  a  resist- 
ance value  on  this  vertical  line  makes  the  chart  adapt- 
able to  resistance  values  corresponding  to  conductors 
at  any  temperature.  Had  the  power  factor  been  lead- 
ing, in  place  of  lagging,  the  corresponding  resistance 
point  would  have  been  located  on  the  upper  half  of  the 
vertical  resistance  line. 

Continuing  following  the  directions  on  the  chart  for 
short  lines,  we  obtain  the  following.  Since  the  circuit 
is  single-phase,  use  2  V  =^  1.24 

,,    ,,  ,      ^  ..  ,     X  c  lOOOOO  X  4000  X    16  X    1.24 

Voltage  drop  in  percent  of  E,  =    2ooocf 

==  19.84  percent 

The  voltage  drop,  as  calculated  mathematically,  is 
19.98  percent  representing  an  error  of  0.14  percent  by 
the  chart. 

Three-Phase  Problem  (No.  jj) — Find  the  voltage 
at  the  sending  end  of  a  three-phase  circuit,  20  miles 
long,  consisting  of  three  No.  ocoo  stranded,  hard-drawn, 
copper  conductors,  spaced  three  feet  apart  in  a  delta 
arrangement.  Temperature  taken  as  25  degrees  C. 
Load  conditions  at  receiving  end  assumed  as  1300  kv-a 
(1040  kw  at  80  percent  power-factor  lagging)  10  000 
volts,  three-phase,  60  cycles. 

From  Table  II,  the  resistance  per  wire  per  mile  is 

again  found  to  be  0.277  ohm  and  since  the  spacing  and 

frequency  are  both  the  same  as  in  the  case  of  the  above 

single-phase  problem,  we  again  obtain  V  =  0.62.     The 

voltage  drop  in  percent  of  £,  is  therefore 

100  000  X  1300  X  20  X  0.62 

, ^1 =  16.12  percent 

10  000 

The  voltage  drop  as  calculated  mathematically  is 

16.16  percent,  representing  an  error  of  0.04  percent. 

CAPACITANCE 

In  long  circuits  the  effect  of  capitance  is  to  de- 
crease the  voltage  drop,  or  increase  the  voltage  rise,  as 


56 


PERFORMANCE  OF  SHORT  TRANSMISSION  LINES 


CHART-IV  DWIGHT  CHART 


FOR  DETERMINING  THE  VOLTAGE  REGULATION  OF  TRANSMISSION  CIRCUITS  CONTAINING  CAPACITANCE 


CORRECT  WITHIN  APPROXIMATELY  ONE  HALF  OF  ONE  PER  CENT  ^  ^0^^  o%^ 

OF  LINE  VOLTAGE  FOR  LINES  UP  TO  100  MILES  LONG  AND  FOR  ,<^V-^  ®    *" 


LOADS  GIVING  NOT  MORE  THAN  15  PER  CENT 
RESISTANCE  OR  REACTANCE  VOLTS  ,  ->« 


^^NO.  3  ALUMINUM 
7  1.7 -p- NO.  6  COPPER  (J- 

o 
^o 

ub 

ll'*J.4-       NO.  2  ALUMINUM       g  i_ 
^NO.  4  COPPER  < 

q: 

LJli. 

u 

Ot3 


I  ALUMINUM 
NO.  3  COPPER 


zz 


^NO.  0  ALUMINUM        Zi  Ld 
"^NO.  2  COPPER  V)  _l 

u 
q: 

NO.  00  ALUMINUM 
NO.  I  COPPER 

NO.  000  ALUMINUM 
Na  0  COPPER 
NO.  0000  ALUMINUM 
NO.  00  COPPER 
250.000  CM  ALUMINUM 
NO.  000  COPPER 
■300,000  CM  ALUMINUM 
NO.  0000  COPPER 
■350.000  CM  ALUMINUM 
'250.000  CM  COPPER 


CO 

IT 


o 

Z 

^       O      <     3 
"S  it!    3 

5  o 
o 


ID 


DIRECTIONS  —Q, 

LAY  A  STRAIGHT  EDGE  ACROSS         T^i 
THE  CHART  FROM  THE  SPACING  POINT 
TO  THE  RESISTANCE  POINT.  AND  READ  V 

SHORT    LI  N  ES 
VOLTAGE  DROP  IN  PER  CENT  OF  E 

100.000  KVA  X  LV 


250.000  CM  COPPER 
1^350000  CM  ALUMINUM 
"^O.  0000  COPPER 
^   300,000  CM  ALUMINUM 
■  —NO.  000  COPPER 
^250,000  CM  ALUMINUM 
NO.  00  COPPER 
NO.  0000  ALUMINUM 

NO.  0  COPPER 
NO.  000  ALUMINUM 

j^NO.  I  COPPER 

^NO.  00  ALUMINUM     J 

^NO.  2  OOPPER  'q 

"^NO.  0  ALUMINUM       ^.  (yj 


WHERE 


DROP  IN  VOLTS  -  1000  KVA  X  LV 

"" E 

KVA  -  KVA  OF  LOAD  AT  RECEIVER  END 
L  -  LENGTH  OF  LINE  IN  MILES 
E  —  LINE  VOLTAGE  AT  LOAD  OR  RECEIVER  END 
V- FACTOR    READ    FROM    CHART.      V    REPRESENTS   A    DROP    IN 
VOLTAGE   WHEN  IT  IS  READ  ON  THE  SAME  SIDE  OF  THE  ZERO 
LINE  AS  THE  RESISTANCE  POINT.  AND  A  RISE  IN  VOLTAGE  WHEN 
IT  IS  ON  THE  OPPOSITE  SIDE. 

LINES  OVER  30  MILES  LONG 
FOR  LONG  LINES.  THE  LINE  CAPACITANCE  DECREASES  THE  DROP,  OR  INCREASES 
THE  RISE   AS  FOUND  ABOVE.  BY  /    ,    \2  /   i     \2 

'°°  ^  (iooo)    ^^^  "^^N"^-  °"  ^'^(iooo)   ^°'-"rS  WHERE 

\'°°"'  V"00/  ^=2.16  FOR  60  CYCLES 

K-0  375  FOR  25  CYCLES 
THE  EFFECTIVE  SPACING  OF  3  PHASE  LINES  IS   S  -^/^^       S  -  1.26  A  FOR  FLAT  SPACING. 

THE  ABOVE  EQUATIONS  ARE  FOR  2  AND  3  PHASE  LINES 

FOR  SINGLE  PHASE  USE  2  V1N  PLACE  OF  V  COPYRIGHT  1916  BY  H.  B.  DWIGHT 


■  ^NO.  3  COPPER  ^      . 

-  *^NO  I  ALUMINUM       [J  li. 


< 

q-q: 

NO.  4  COPPER         2  _ 
NO.  2  ALUMINUM     ^  jj 

l-O 
(0  < 

u 

rNO.  6  COPPER 
NO.  3  ALUMINUM 

RESISTANCE-OHMS 
ONE  CONDUCTOR 


PERFORMANCE  OF  SHORT  TRANSMISSION  LINES 


S7 


will  be  explained  later.  The  Dwight  and  Mershon 
charts  do  not  recognize  the  effect  which  capacitance  has 
upon  the  voltage  drop.  In  the  lower  left  hand  corner 
of  the  Dwight  Chart,  however,  there  is  placed  a  formula 
by  which  a  correction  may  be  applied  to  the  voltage 
drop  as  given  by  the  chart  This  correction  accounts 
for  the  effect  of  the  charging  current  (resulting  from 
capacitance)  quite  accurately,  provided  the  circuit  is 
not  too  long  or  the  frequency  too  high.  The  application 
of  this  corrective  factor  will  be  evident  from  the  follow- 
ing problem. 

TABLE  K-COSINES,  SINES  AND  TANGENTS 


ANGLE 

cos  0 

(P  F) 

SIN  8 

TAN  0 

0°  00' 

1. 000 

0.0000 

0.0000 

8'    o6' 

0.990 

0.1409 

0.1423 

11°   28' 

0.980 

0.1988 

0.2028 

'^o  °4' 

0.970 

0.2430 

0.2506 

't    '5 

0.960 

0.2798 

O.29IS 

i8°  n' 

0.950 

0.3120 

0.3285 

'<   56; 

0.940 

0.3410 

0.3627 

21°  33 

0.930 

0.3673 

0.3949 

<    04 

0.920 

0.3918 

0.4258 

24:  ^ 

0.910 

0.4144 

0.4554 

25„  50' 

0.900 

0.4357 

0.4841 

^L  °7' 

0.890 

0.4558 

O.512I 

28°  21' 

0.880 

0.4748 

0.5396 

29^  32' 

0.870 

0.4929 

0.5665 

30„  41 

0.860 

0.5103 

0.5934 

31°  47' 

0.850 

0.5267 

0.6196 

3^0  si; 

0.840 

0.5424 

0.6457 

33   54 

0.830 

0.5577 

0.6720 

34°  54' 

0.820 

0.5721 

0.6976 

35   54' 

0.810 

0.5864 

0.7239 

36   52; 

0.800 

0.6000 

0.7499 

37°  48' 

0.790 

0.6129 

0.7757 

^K    44' 

0.780 

0.6257 

0.8021 

39°  38 

0.770 

0.6379 

0.8283 

40   32 

0.760 

0.6499 

0.8551 

K     ^ 

0.750 

0.6613 

0.8816 

42°  16' 

0.740 

0.6726 

0.9089 

43°  06' 

0.730 

0.6833 

0.9358 

43°  56' 

0.720 

0.6938 

0.9634 

44,  45; 

0.710 

0.7040 

0.9913 

"•1°  ^ 

0.700 

0.7141 

1.0199 

46°  22' 

0.690 

0.7238 

1.0489 

47°  09' 

0.680 

0.7331 

1.0780 

47   ss; 

0.670 

0.7422 

I.IO74 

<  42 

0.660 

0.7513 

1.1383 

<  27 

0.650 

0.7598 

1.1688 

50°   12' 

0.640 

0.7683 

1.2002 

50°  57' 

0.630 

0.7766 

1.2327 

5'„°  41; 

0.620 

0.7846 

J.26S5 

52°  24' 

0.610 

0.7923 

1.2985 

53:  °7' 

0.600 

0,8000 

1.3327 

S3   SO 

0.590 

0.8073 

1.3680 

54   32' 

0.580 

0.8145 

'■4037 

55   14' 

0.570 

0.8215 

1.4406 

55°  56' 

0.560 

0.8284 

1.4788 

56°  37' 

0550 

0.8350 

1.5175 

57   18' 

0.540 

0.8415 

1.5577 

57°  59' 

0.530 

0.8479 

1.5993 

58   40; 

0.520 

0.8542 

1.6426 

59   20' 

0.510 

0,8601 

1.6864 

60°  00' 

0.500 

0.8660 

1.7320 

60°  39' 

0.490 

0.8716 

1.7783 

61°  18' 

0.480 

0.8771 

1.8265 

61°  57' 

0.470 

0.8825 

1.8768 

Three-Phase  Problem  (No.  45) — Find  the  volt- 
age at  the  sending  end  of  a  three-phase  circuit,  100 
miles  long,  consisting  of  three  No.  0000,  stranded,  hard- 
drawn  copper  conductors,  spaced  nine  feet  apart  in  a 
delta  arrangement.  Temperature  assumed  as  25  de- 
grees C.  Load  conditions  at  receiving  end  assumed  as 
22000  kv-a,  80  percent  power-factor  lagging,  88chx) 
volts,  60  cycles. 


load   by    the    amount   of    100  X  ^- 


.y 


=   2.l6 


From  Table  II  the  resistance  is  found  to  be  0.277 

ohm  per  mile.      From  Dwight  Chart  read  V  =  0.70. 

Then,  the  voltage  drop  in  percent  of  £„  if  the  line  were 

short,  would  be, 

looooo  X  22000  X  100  X  0.70 

83ooo» =  19.89  percent 

From  directions  on  the  Dwight  chart  for  circuits 
over  30  miles  long,  the  charging  current  of  this  circuit 
is  found  to  be  such  as  to  decrease  the  voltage  drop  un- 
der load  conditions  or  to  increase  the  voltage  at  zero 

\/OOOj 

percent.  Hence  the  voltage  at  the  sending  end,  under 
load  conditions,  will  be  ip.8p  —  2.16  ^^^  17-73  percent. 
The  actual  result  as  calculated  rigorously  is  17.94  per- 
cent. Thus  the  error  by  the  Dwight  graphical  solution 
is  approximately  0.21  percent. 

If  the  power-factor  of  the  load  is  assumed  as  100 
percent  (problem  46)  in  place  of  80  percent  lagging,  we 
get  V  =  0.3s  and  find  the  error  for  the  Dwight  graphi- 
cal solution  of  this  100  mile,  60  cycle  circuit  to  be  ap- 
proximately 0.75  percent.  It  should  be  noted,  however, 
that  the  reactance  volts  are  in  this  case  22  percent  of 
the  receiving  end  voltage. 

SENDING  END  CONDITIONS  FIXED 

When  the  sending  end  conditions  are  fixed,  a  dif- 
ferent form  of  solution  must  be  employed  to  determine 
the  size  of  conductors  corresponding  to  a  given  voltage 
drop.  In  such  cases,  the  Dwight  Chart  is  particularly 
applicable.  To  use  the  chart  for  the  solution  of  such 
problems  proceed  as  follows.  First  V  is  calculated  by 
means  of  the  formulas  on  the  chart,  and  then  a  straight 
edge  is  placed  through  V  (on  the  line  corresponding  to 
the  power-factor  of  the  load)  and  the  point  for  the 
spacing  and  frequency  to  be  used,  and  the  required  size 
of  conductor  can  be  seen  at  a  glance  on  the  resistance 
scale  at  the  right.  To  make  this  application  of  the 
chart  clear,  the  following  is  given, — 


Voltage  drop  in  percent  of  Er  ■ 
Hence 


100  000  Kv-a  y.  L  V 


£,' 


iiS) 


V  = 


Voltage  drop  in  percent  of  Et  X  E* 


(^9) 


100  000  Kv-a  XL  

Applying  (29)  to  the  above  problem  No.  33  we  get 

16.12  X  10  000'        _ 
~iooooo  X  1300  X  20    ~  **■  ^ 
Following  the  above  directions,  the  resistance  per 
mile  is  found  to  be  0.277  ohm  and  the  corresponding 
size  of  conductor  No.  0000  copper. 

MATHEMATICAL  SOLUTION 

In  order  to  check  any  one,  or  all  of  the  above  de- 
scribed graphical  methods,  a  complete  mathematical 
solution  may  be  made  by  applying  the  various  trigono- 
metrical formulas.  Fig.  18,  to  the  values  of  the  problem 
under  consideration.  These  formulas  have  been  ar- 
ranged to  meet  the  conditions  of  loads  of  either  lagging 
or  leading  power-factors,  and  for  conditions  fixed  at 
either  the  receiving  or  the  senuing  ends. 

There  are  numerous  problems  requiring  a  solution 


S8 


PERFORMANCE  OF  SHORT  TRANSMISSION  LINES 


where  the  voltage  at  the  sending  end,  and  the  kilowatts 
and  the  power-factor  of  the  load  at  the  receiving  end 
are  fixed.  In  such  cases  it  is  required  to  determine  the 
corresponding  receiving  end  voltage.  This  determina- 
tion can  be  made  mathematically,  but  such  a  solution  is 
tedious,  since  the  formulas  applying  to  such  cases  are 
cumbersome.  Formulas  are  given  at  the  bottom  of  Fig. 
i8  which  may  be  applied  to  such  problems.  Time  and 
labor  may,  however,  be  saved  in  solving  such  problems 
by  the  employment  of  a  cut-and-try  method  usually  used 
in  such  cases,  as  follows: — 

Assume  what  the  voltage  drop  will  be,  correspond- 
ing to  the  size  of  conductors  likely  to  be  used.  On  the 
basis  of  this  assumption  the  receiving  end  voltage  is 
fixed;  thus,  all  of  the  receiving  end  conditions  are  as- 
sumed to  be  fixed.  The  corresponding  sending  end 
voltage  is  then  readily  determined  by  one  of  the  graphi- 
cal methods  described.  If  the  sending  end  voltage  thus 
determined  is  found  to  be  materially  different  from  the 
fixed  sending  end  voltage,  another  trial,  based  upon  a 
different  receiving  end  voltage,  will  probably  suffice. 

Single-Phase  Problem — Find  the  characteristics  of 
the  load  at  the  sending  end  of  a  single-phase  circuit, 
16  miles  long,  consisting  of  two  stranded,  hard  drawn, 
copper  conductors,  spaced  three  feet  apart ;  temperature 
taken  as  25  degrees  C  ;  load  conditions  at  receiving  end 
assumed  as  4000  kv-a  (3200  kw  at  80  percent  power- 
factor  lagging)  20000  volts,  60  cycles;  transmission 
loss  to  be  approximately  ten  percent. 

Following  the  procedure  given  in  Chart  II,  consult 
Quick  Estimating  Table  XVII  for  a  delivered  voltage 
of  20000.  Since  the  conditions  of  the  above  problem 
are  a  power-factor  of  80  percent,  and  a  temperature  25 
degrees  C,  the  corresponding  kv-a  values  are  as  in- 
dicated at  the  head  of  the  table  on  the  basis  of  10.8  per- 
cent loss  in  transmission  for  a  three-phase  circuit.  For 
a  single-phase  circuit  the  corresponding  values  will  be 
one-half  the  table  values.  Thus  the  4000  kv-a  single 
phase  circuit  of  the  problem  is  equivalent  to  8000  kv-a, 
three-phase  on  the  table.  From  the  table,  it  is  seen  that 
for  a  distance  of  16  miles  7810  kv-a,  three-phase  can 
be  transmitted  over  No.  0000  conductors  with  a  loss  of 
10.8  percent.  7810  kv-a  is  near  enough  to  8000  kv-a,  and 
the  loss  of  10.8  percent  is  near  enough  to  an  assumed 
loss  of  ten  percent,  so  we  decide  that  No.  0000  copper 
•  conductors  come  nearest  to  the  proper  size  to  meet  the 
conditions  of  the  problem.     The  loss  with  No.  0000 

8000 
conductors  will  be-Q-   X  10.8  =11.06  percent,  as  will 

be  shown  later. 

Table  XXIII  indicates  that  there  will  be  no  over- 
heating of  this  size  of  conductor. 

Table  XXII  indicates  that  20  000  volts  is  too  low  to 

result  in  corona  loss  with  No.  0000  conductors,  at  any 

reasonable  altitude.     Then, — 

V  4000 

Kv-a,n   :=  — ^  2000  kv-a  to  neutral. 

3200 
Kwrm  =  — — ^=  1600  kw  to  neutral. 


Ei,  =  =  10  000  volts  to  neutral. 

2000000  .     . 

'  ^  '  10000     =  200  amperes  per  conductor. 

The  resistance  per  conductor  is 

i?  =  16  X  0-277  (from  Table  II)  =  4.432  ohms. 

The  reactance  per  conductor  is 

X  =  16  X  0.644  (from  Table  V)  =  10.304  ohms. 
and  IR  =  200  X  4-432  =  866  volts,  resistance  drop 
886 

=  X  100  =:  8.86  percent 

10  000  ^  '^ 

IX  =  200  X  10.304  =  2061  volts,  reactance  drop 

= X  100  =  20.61  percent 

10  000     'j 

E.n  =  1/(10000  X  0.8  -I-  866)'  4-  (loooo  X  0.6  +  2061)' 

=  II 998  volts  to  neutral (30) 

,/ (10  000  X  0-6) +2061  \ 

«•  =''^''-'1(10  000X0.8) -f  886   J  =  ^2    13     (jO 

Percent  PF.  =  (Cos.  42°  13' )  X  100  =  7406  percent (3s) 

Kv-a,a  =  ^°°  10^ =  2399.6  kv-a  per  conductor (33) 

Kw,n  =  2399.6  X  0.7406  =  1777-1  kw  per  conductor (34) 

11998— lOOOO  ^^  o  ..  . 

Percent  voltage  drop  = ^^^ X  100  =  19-98  percent 

(46) 

Transmission  loss  =  ^^^^  "^^^  =  '77-28  k^  Per  conductor 
(47) 

177  28  ^^  2 

Percent  transmission  loss  = "     pp        X  loo  =  11.08  percent 

(4S) 

Three-Phase  Problem  (No.  55)— Find  the  char- 
acteristics of  the  load  at  the  sending  end  of  a  three- 
phase  circuit  20  miles  long,  consisting  of  three  stranded, 
hard-drawn,  copper  conductors,  spaced  in  a  three  foot 
delta.  Temperature  taken  as  25  degrees  C.  Load 
conditions  at  receiving  end  assumed  as  1300  kv-a.  (1040 
kw  at  80  percent  power-factor  lagging)  10  000  volts,  60 
cycles ;  transmission  loss  not  to  exceed  ten  percent. 

Following  the  procedure  given  in  Chart  II,  the  fol- 
lowing results  are  obtained : — 

Consult  Table  XV  for  a  delivered  voltage  of  10  000 
volts.  Since  the  conditions  of  the  above  problems  are, 
power-factor  of  load  80  percent,  temperature  25  degrees 
C.  the  corresponding  three-phase  kv-a  values  of  the 
table  are  on  the  basis  of  10.8  percent  loss  in  transmis- 
sion. From  Table  XV  it  is  seen  that  1240  kv-a,  three- 
phase  can  be  transmitted  over  No.  000  conductors,  or 
15(30  kv-a.,  three-phase  over  No.  0000  conductors  at 
10.8  percent  loss.  Since  the  loss  for  the  problem  is 
not  to  exceed  ten  percent  and  1300  kv-a  is  to  be  tnms- 
mitted,  we  will  select  No.  0000  conductors.     The  loss 

j-jOO 

for  these  conductors  will  therefore  be  — —-  of  10.8.  or 

1560 

nine  percent  as  will  be  shown  later. 

Table  XXIII  indicates  that  there  will  be  no  over- 
heating of  this  size  of  conductor  when  carrying  1300 
kv-a,  three-phase. 

Table  XXII  indicates  that  10000  volts  Is  too  low 
to  result  in  corona  loss  with  No.  OOGG  conductors  at 
any  reasonable  altitude.     Then: — 

J-3QQ 

Kv-On  = — I —   =  433.33  kv-a  to  neutral. 
Kwt,  — — —  =  346.6  kw  to  neutral. 


and 


_  lOOOO 

'■  ~  r.732  ~  5774  volts  to  neutral. 

,  433  333 

'r  =  —  =  75.05  amperes  per  conductor. 

The  resistance  per  conductor  is, — 

i?  =  20  X  0.277  (from  Table  II)  =  5.54  ohms. 

The  reactance  per  conductor  is, — 

X  =  20  X  0.644  (from  Table  V)  =  12.88  ohms. 

IR  =  75.05  X  5-54  =  415-8  volts,  resistance  drop. 
4158 


PERFORMANCE  OF  SHORT  TRANSMISSION  LINES  59 

Transmission  loss  =     ^^"^^     ^^=  3120  *w  per  conductor 

(^) 

9.00  percent 


%t  20  X  1 

Percent  transmission  loss  =  -^^-^ — ^^"^X  lOO  = 

1040 


S774 


X  100  =  7.20  percent. 


OIAORAM 

FOR 

LAOQINQ  POWEH-FAOTOR 


MIXED  SENDING  AND  RECEIVING  END  CONDITIONS  FIXED 

Brancn  circuits  are  frequently  run  from  the  main 
transmission  trunk  circuit  to  the  center  of  some  local 
distribution.  In  such  cases  the  voltage  at  the  sending 
end  and  the  current  or  the  power  and  power-factor  at 


-^8 


••CURIJENT 


DIAGRAM 

FOR 

LEADING  J>OWER-F ACTOR 


P^ 


•    V       i      '  i 

_i_i i h-jL* 

K. ERCOsep  — -»i        ! 

V EgCOSBg. »J 


CURRENT 


LOADS  OF  LAGGING   POWER- FACTOR 


^^ 


LOADS  OF  LEADING   POWER -FACTOR 


£3=  y(ER  cos  Sr+IR)^  +  (Er  sin  Sh  +  OCf        (30) 
(ErSIN9r)+1X) 

WHEN  es=TAN-l     ,EoCOSeH^+IR)       "'* 

RECEIVING-END  V  "  ")        f 

CONDITIONS       %  PFs=-  COS  Og  X  100       (32) 
ARE  FIXED 

KVAcu-LlrSN  PER  CONDUCTOR       (33) 
^'^    "iooo 
KW^K|-  KV-Aqki  X  COS  9<~  PER  CONDUCTOR       (34) 


Er-  ^(EsCOS6s-IR/  +  (EgSINeg-lX)^       (36) 


(Eg  SIN  Sg)- IX) 

WHEN  Sr-TAN-I     /EgCOS9s)-IR)      "" 

SENDING-ENO  ^  '"  "11 

CONDITIONS       %  PFr-  COS  Or  X 100      (37) 
ARE  FIXED  ,^^ 

KVAo.,"         BH  PER  CONDUCTOR       (38) 


"RN" 


1000 


KWrm-KVArnX  COS  Sr  per  CONDUCTOR       (39) 


Eg-  "^CEr  COS  eR+  1R)^  +  (Er  SIN  9r  -  IX)  (40) 

(ERS.NeR)-DC) 
WHEN  Sg-TAN-*    (Er  COS  9r)+1R)      '*" 

RECEIVING -END 
CONDITIONS       %PFg-OOS9sXl00       (32) 

ARE  FIXED        ^u.  l«EsN  pER  CONDUCTOR       (33) 

SN       IOOO 
KW;„-  KVAgfi  X  COS  9^  PER  CONDUCTOR       (34) 


Er-  y(Es  COS  98  -  IR)^  +  (Eg  SIN  9s-t-lX)^        (4J) 
(EgSINBs)-t-D<) 

WHEN  On-TAN-<     /Vc  C03Bc\-IR->      <*» 

SENDING-ENO  "  ('^S'-^'S)    '") 

CONDITIONS        «  pp   _  COS  9r  X  100       (37) 
ARE  FIXED  "  " 

in/.An,i— '"^RN  PER  CONDUCTOR      (38) 
""       IOOO 
kWp,,-  KVAnfi  X  COS  Br  PER  CONDUCTOR       (39) 


GENERAL  FORMULAS 


WHEN  THE  VOLTAGE  AT  SENDING  END  AND  THE  AMPERES  AND  POWER^^ACTOR  AT  RECEIVINOEND  ARE  FIXED 


Er  =  - 1  (r  COS  9r  ±X  SIN  eR)+  VeI-i'  (r^  SIN^  Sr+X^  C0S°  9r)±  2 1^  R  X  COS  9r  SIN  8r    (44) 
*  USE  +  WHEN  THE  POWER-FACTOR  OF  THE  LOAD  IS  LAGGING  AND  -  WHEN  THE  POWER-FACTOR  IS  LEADING 


WHEN  THE  VOLTAGE  AT  SENDING  END  AND  THE  POWER  AND  POWER-FACTOR  AT  RECEIVING  END  ARE  FIXED 
(POWER  FACTOR  LAGGING) 


j  =  A  Wit  Vl- 


(RnX')KWRN^XI08 
A*  COS*  Or 


AT,      1 000  KWru  (R  cos  9o  +  X  SIN  9r) 
WHEREA-Eg^yX eilL__a^ !    ,4« 


i2r 


%  VOLTAGE  DR0P= 


-S~'-R 


XIOO      (46) 


TRANSMISSION  LOSS-  j^  KW  PER  CONDUCTOR       (47  > 
%  TRANSMISSION  LOSS-  -°tOTAl"kWb''*"  ^ '  °°       ***' 


FIG.    l8 — TRICWNOMETRICAL    FORMULAS    FOR    SHORT     TRANSMISSION   UNES 

Capacitance  effect  not  taken  into  account. 


IX  =  75.05  X  12.88  =  966.6  volts,  reactance  drop. 
966.6   ,, 
=  ■        -  X  100  =  16.74  percent. 


the   receiving  end  are  approximately  fixed.     In   such 

cases  the  calculation  for  the  voltage  at  the  receiving  end 

, — ^         -,^,   ,    ,    ,   ,    ,^,,,    ,  requires  more  arithmetical  work  than  is  required  when 

£..  =  1/  (5774  X  0.8 -f- 415-8)' -h  (5744  X  0.6 -f966.6  "=6707  ,,'.  ...  ^  a      t  .x,        ■        ■.•  c      a 

volts  to  neutral (30)  all  the  conditions  at  one  end  of  the  circuit  are  fi.xed. 

(5774  X  0.6  -f  966.6  \  _     o      ,  f    ,,  Such  problems  can  be  more  readily  solved  graphically, 

S774X0.8  4-41S-8  /  ~  ^  "  as   previously   explained,   but    may   be   solved   mathe- 

PF.  =  (Cos  41"  22')  X  100  =  75-05  percent (3^)  matically  by  applying  formula  (44)  or  (45),  Fig.  18. 

Kv-a„  =  —j~ — ^^^  —  S03.4  kv-a  per  conductor (33)  To  illustrate  the  application  of  formula   (44)   we 

Kw,.  =  503.4  X  0.7505  =  377-8  kw  per  conductor (3^)  will  apply  the  values  of  Problem  jj  to  formula  (44)  and 

Percent  voltage  drop  =-^^^^r=^  X  loo  =  16.16  percent  calculate  the  receiving  end  voltage.     Thus  we  have  as 


S774 


(.46) 


fixed  conditions: — 


6o 


PERFORMANCE  OF  SHORT  TRANSMISSION  LINES 


En   =  6707  volts 

It  =  75-05  amperes 
■Cos  0,  =  0.8 
Sin  S,  =:  0.6 

R  ===^  5-54  ohms 
X  =  12.88  ohms 
IR  =  415.8  volts 
Then 

£r  =  —7505  (5-54  X  0.8  +  i2.{ 


X  0.6)  + 


1/6707'  —  75-05'  (5-54'  X  0.6-'  +  12.88=  X  0.8')  +  2  X 

7505"  X  5-54  X  12.88  X  0.8  X  0.6 U4) 

=  —  913  +  F    44983849  —  660242  +  385831 
=  —  913  +  6637  =  5774  volts. 
To  illustrate  the  application  of  formula  (45)  we 
v/ill  apply  the  values  of  Problem  55  to  formula  (45) 

TABLE   L 
ILLUSTRATING  VARIATION    IN    REACTANCE 

Resulting  from  Changes  in  the  Conductors  and  Transmission  Voltages 


and  calculate  the  receiving  end  voltage.     Thus  we  have 
as  fixed  conditions : — 
E.n  =  6707  volts   ■ 
Kw,„  =  346.6  kw 
R  =  5.54  ohms 
X  =^  12.88  ohms 

Cos  6,  ■=:  0.8 

Sin  Br  =  0.6 


then 


A  =  6707 -y/  0.5 


loooX  346.6  (5.54  X  0.8  +  12.88  X  0.6) 


6707'  X  0.8 


E,.  =  AJ^+y 


i 


A 


(5-54'  +  12.88')  346.6'  X  iC 
A'  X  0.8' 


Us) 


-  6707  j/'o.s  —  0.1172  =  4152 


=  41521/  I  +  0.936  =:  5774  volts 

Alternative  to  (44)  and  (^5)— The  following 
formulas  have  been  proposed  by  Mr.  H.  B.  Dwight  to 
meet  the  mixed  conditions  referred  to, — 

E,„  :=  6707  volts 

1000  X  Kwr,  =  346600  wafts 

1000  X  reactive  Kv-a,^  =  346600   X  -^=  260  000  v-a 

R  =  5.54  ohms 

X  =  12.88  ohms 

L  =  346600  X  5-54  +  260000  X  12.88  =  5270000 

M  =  346600  X  12.88  —  260000  X  5-54  =  3025000 


E^  =  0.5  E^^  -  L+  o.s  VE,*  -  y  E,^  I.  -  4  M"- 
•E  =  5774  volts 


R  -  R       1-      £.       J^ 

^  -  ^'~  E,-  E,'~  ^E,^' 


Ei' 


2      E,' 


5L^ 
E,- 


~      E,'      - 
E  =  5779  volts 


E,' 


CONDUCTORS 

Total 
I'  R 

Loss 
(KW) 

IR 

IX 

Approximate 

Voltage 
Regulation  at 

Volts 

Per 
Cent. 

Volts 

Per 
Cent. 

100 

Per 

Cent. 

Power 

Factor 

80  Per 
Cent. 
Power 
Factor 
(Lag.) 

RECEIVING  END   VOLTAGE  —  0600 

Single    Circuit    of 
three  500,000  circ.mil 
bare  overhead  con- 
ductors 

129 

123 

3.22 

622 

16.32 

4.5 

12.8 

Two  circuits  each 
of  three  250,000  circ. 
mil    bare    overhead 
conductors. 

129 

123 

3.22 

133 

8.73 

3.< 

7.7 

One      Circuit      of 
500,000     circ.      mil 
three-conductor 
cable.        Insulation 
thickness    ij   by   a 
inches. 

129 

1.23 

3.22 

172 

4.52 

3.2 

5.0 

RECEIVING  END  VOLTAGE  —  13  200 

Single     circuit    of 
three     126,000     circ. 
mil    bare    overhead 
conductors. 

129 

247 

3.22 

354 

4.M 

3.2 

5.1 

CIRCUITS    OF    EXCESSIVE    REACTANCE 

If  a  large  amount  of  power  is  to  be  transmitted  at 
comparatively  low  voltage,  particularly  if  the  frequency 
is  high,  the  reactance  of  the  circuit  will  be  high  com- 
pared with  its  resistance.  If  the  reactance  is  excessive 
(20  to  30  percent  reactance  volts  may  in  some  cases  be 
considered  excessive),  the  voltage  regulation  of  the  cir- 
cuit may  be  seriously  impaired. 

As  will  be  seen  by  consulting  Tables  VI  and  VII, 
there  is  a  fixed  relation  between  tlie  resistance  and  the 
reactance  of  a  circuit  for  a  given  frequency,  size  and 
spacing  of  conductors.  This  ratio  is  2.4  times  greater 
for  60  cycle  than  it  is  for  25  cycle  circuits.  For  a  given 
size  of  conductor  the  reactance  can  be  varied  only 
slightly  by  changing  the  spacing  of  overhead  bare  con- 
ductors. Substituting  a  larger  or  smaller  conductor 
may  change  the  resistance  materially,  but  this  will  have 
little  effect  upon  the  reactance. 

The  reactance  may  be  reduced  by  either  or  all  of 
the  following  methods.  The  circuit  may  be  split  up 
into  two  or  more  circuits  employing  smaller  conduc- 
tors and  these  circuits  connected  in  parallel.  The  volt- 
age may  be  raised,  if  the  installation  is  new,  and  smaller 
conductors  employed;  or  the  overhead  conductors  may 
be  replaced  by  three  conductor  cables.  To  illustrate 
the  above  methods,  the  following  problem  has  been  as- 
sumed and  the  results  tabulated. 

A  HIGH  REACTANCE  PROBLEM 

Table  L  refers  to  the  following  problem — ^,}.ooo 
kv-a,  three-phase,  60  cycles,  is  to  be  delivered  a  dis- 
tance of  three  miles  over  hard-drawn,  stranded  copper 
conductors.  The  PR  loss  is  to  remain  at  129  kw.  The 
spacing  of  the  overhead  conductors  assumed  as  3  by  3 
by  3  ft.     Temperature  25  degrees  C. 

It  is  evident  from  Table  L  that  if  two  three-phase 
circuits,  each  consisting  of  three  250000  circ.  mil.  con- 
ductors are  installed  in  place  of  one  three-phase  circuit, 
consisting  of  three  500  000  circ.  mil.  conductors,  the  re- 
actance will  be  reduced  by  nearly  one  half,  and  a  cor- 
responding improvement  in  the  voltage  drop  or  regula- 
tion will  occur,  particularly  if  the  load  power-factor  is 
80  percent  lagging.  A  further  improvement  along  this 
line  will  be  obtained  if  a  single  three-conductor  cable  is 
employed.  Doubling  the  voltage  for  the  overhead  cir- 
cuit and  employing  three  125  000  circ.  mil.  conductors 
results  in  practically  as  good  performance  in  voltage 
regulation  as  for  the  6600  volt  three-conductor  cable. 


*See  article  by  Mr.  H.  B.  Dwight  on  "Effect  of  a  Tie  Line 
between  Two  Substations"  in  the  Electrical  Review,  Dec.  21, 
1918,  p.  966.  The  formulas  given  in  this  article  make  complete 
allowance  for  the  effect  of  capacitance  and  are  very  similar  to 
the  above. 


CHAPTER    VIII 
PERFORMANCE  OF  LONG  TRANSMISSION  LINES 

(GRAPHICAL  SOLUTION) 


THE  E.M.F.  of  self-induction  in  a  transmission 
circuit  may  either  add  to  or  subtract  from  the 
impressed  voltage  at  the  sending  end,  depend- 
ing upon  the  relative  phase  relations  between  the  cur- 
rent and  the  voltage  at  the  receiving  end  of  the  circuit. 
This  is  illustrated  by  means  of  voltage  vectors  in  Fig. 
20,  in  which  the  phase  of  the  current  is  assumed  to  be 
constant  in  the  horizontal  direction  indicated  by  the 
arrow  on  the  end  of  the  current  vector.  The  voltage  at 
the  receiving  end  is  also  assumed  as  constant  at  100 
volts.  The  vector  representing  the  receiving  end  volt- 
age (Er  =  100  volts)  is  shown  in  two  positions  corre- 
sponding to  leading  current,  two  positions  correspond- 
ing to  lagging  current  and  in  one  position  corresponding 
to  unity  power-factor.  The  components  IR  and  IX  of 
the  supply  voltage  necessary  to  overcome  the  resistance 
R  and  the  reactance  X  (e.m.f.  of  self-induction)  of  the 
circuit  are  assumed  to  be  10  volts  and  20  volts  respec- 
tively. Since  the  current  is  assumed  as  constant.  IX 
and  IR  are  also  constant.  The  impedance  triangle  of 
the  voltage  components  required  to  overcome  the  com- 
bined effect  of  the  resistance  and  the  reactance  of  this 
circuit  is  therefore  constant.  It  is  shown  in  five  differ- 
ent positions  about  the  semicircle,  corresponding  to  five 
different  load  power-factors.  The  voltage  E,  at  the 
sending-end  required  to  maintain  100  volts  at  the  re- 
ceiving-end is  indicated  for  each  of  the  five  positions  of 
the  impedance  triangle. 

Counter-clockwise  rotation  of  the  vectors  will  be 
considered  as  positive.  This  means  that  when  th:  cur- 
rent is  lagging  behind  the  impressed  e.m.f.,  the  voltage 
vector  will  be  in  the  forward  or  leading  direction  from 
the  current  vector  as  indicated  by  the  arrow.  When 
the  current  leads  the  impressed  voltage,  the  voltage 
vector  will  be  in  the  opposite,  or  clockwise  direction 
from  the  current  vector.  In  other  words,  assuming  the 
vectors  all  rotating  at  the  same  speed  about  the  poin,  O 
in  a  counter-clockwise  direction,  the  current  vector  will 
be  behind  the  voltage  vector  when  the  current  is  lagging 
and  ahead  of  it  when  the  current  is  leading. 

The  alternating  magnetic  flux  surrounding  the  con- 
ductors, resulting  from  current  flowing  through  them, 
generates  in  them  a  counter  e.m.f.  of  self-induction. 
This  e.m.f.  of  self-induction  has  its  maximum  value 
when  the  current  is  passing  through  zero  and  is  there- 
fore in  lagging  quadrature  with  the  current.  On  the 
c'iagrams  an  arrow  in  the  line  IX,  indicates  the  direction 
of  the  e.m.f.  of  self-induction.  It  will  be  seen  that  since 
the  direction  of  the  current  is  assumed  constant, 
the    e.m.f.    of    self-induction    acts    downward    in    all 


five  impedance  diagrams.  The  sending-end  voltage 
is  therefore  opposed  or  favored  by  this  self- 
induced  voltage  (see  arrows)  to  a  greater  or  less 
extent  depending  upon  the  power-factor  of  the  load. 
Thus  at  lagging  loads  of  high  power-factor,  the  self-in- 
duced voltage  acts  approximately  at  right  angles  to  the 
5ending-end  voltage,  and  therefore  requires  a  small  com- 
ponent of  the  sending-end  voltage  to  balance  or  neutral- 
ize its  effect.  As  the  power-factor  of  the  receiving-end 
load  decreases  in  the  lagging  direction  (upper  quadrant 
of  diagram)  the  sending-end  voltage  swings  around 
more  nearly  in  line  with  the  direction  of  the  induced 
voltage,  thus  requiring  a  greater  component  of  the  send- 
ing-end voltage  to  counter-balance  its  effect.  At  zero 
power-factor  lagging,  the  direction  of  the  sending-end 
voltage  and  that  of  the  induced  e.m.f.  are  practically  in 
opposition,  (as  indicated  by  the  arrows),  so  that  the 
component  of  the  sending-end  voltage  required  to  over- 
come the  induced  voltage  is  a  maximum,  or  nearly  as 
much  as  the  e.m.f.  of  self-induction.  It  is  interesting  to 
note  that  at  zero  lagging  power-factor,  when  the  effect 
of  self-induction  on  line  voltage  drop  reaches  a  maxi- 
mum, the  sending-end  voltage  component  IR  necessary 
to  overcome  the  resistance  of  the  circuit,  (now  nearly 
at  right  angles  to  the  supply  voltage),  is  a  minimum. 
The  reverse  of  these  conditions  is  true  for  receiv'ng- 
end  loads  of  power-factors  near  unity. 

Now  consider  receiving-end  loads  of  leading 
power-factors,  (lower  quadrant  of  diagram).  It  will 
be  seen  that  the  e.m.f.  of  self-induction  does  not  now 
oppose  the  sending-end  voltage  (indicated  by  direction 
of  the  arrows)  but  has  a  direction  more  or  less  parallel 
to  that  of  the  sending-end  voltage.  At  high  leading 
power-factors,  the  e.m.f.  of  self-induction  has  little 
effect  on  the  sending-end  voltage,  but  as  zero  leading 
power-factor  is  approached  these  two  e.m.f.'s  more 
rearly  come  in  phase  with  each  other.  At  zero  power- 
factor  leading,  the  e.m.f.  of  self-induction  adds  almost 
directly  to  the  sending-end  voltage. 

It  will  be  seen,  therefore,  that  for  receiving-end 
loads  of  lagging  power-factor,  the  sending-end  voltage 
is  greater  than  the  receiving-end  voltage,  by  an  amount 
necessary  to  overcome  the  resistance  and  self-induction 
of  the  circuit.  For  receiving-end  loads  of  leading 
power-factor,  the  sending-end  voltage  is  less  than  the 
receiving-end  voltage,  for  the  reason  that  the  e.m.f.  of 
self-induction  is  in  such  a  position  as  to  assist  the  send- 
ing-end voltage. 

The  following  values  from  Fig.  20  illustrate  these 
conditions : 


6a 


GRAPHICAL   SOLUTION  FOR  LONG  LIN  EH 


Power-Factor  of 
Receiving  End  Load 

Supply  Voltage 

0  percent  lagging 
80  percent  lagging 
ICX)  percent 
80  percent  leading 

0  percent  leading 

120.4 
120.4 
111.8 

98.5 
80.6 

The  condition  of  leading  power-factor  at  the  re- 
ceiving-end would  be  unusual  in  practice,  since  the 
power-factor  of  receiving-end  loads  is  usually  lagging. 
In  cases,  however,  where  condensers  are  used  for 
voltage  or  power-factor  control,  the  power-factor  at  the 
receiving-end  may  be  leading.  If  the  circuit  were  with- 
out inductance,  there  could  be  no  rise  in  voltage  at  the 

'■—M'^OlSG^ 


■^P.F  O^ LCtf D  80%  LAGGING 


P.F  OF  LOAD.  80%  LEADING 

BASIS  OF   DIAGRAM 
tr   HELD  CONSTANT  AT  100  VOLTS. 
IR    RESISTANCE  VOLTS  CONSTANT  AT  10  VOLTS. 
IX    REACTANCE  VOLTS  CONSTANT  AT  20  VOLTS. 
.IZ    IMPEDANCE  VOLTS  CONSTANT  AT  22.4  VOLTS, 
-P.F.  OF  LOAD.  0%  LEADING 

FIG.    20 — EFFECT  OF  SELF   INDUCTION  ON   REGULATION 

receiving-end,  for  in  such  a  case,  IX  of  the  diagram 
would  disappear,  and  the  voltage  drop  would  be  the 
same  as  with  direct  current.  All  alternating-current 
circuits  are  inductive,  and  the  greater  their  inductance, 
the  greater  will  be  the  voltage  drop,  or  the  voltage  lise 
along  the  circuit. 

Any  alternating-current  circuit  may  be  looked  upon 
as  containing  three  active  e.m.f.'s  out  of  phase  with  each 
other.  In  addition  to  the  impressed  e.m.f.  at  the  send- 
ing-end,  there  are  two  e.m.f  's  of  self-induction,  one  as 
the  result  of  the  receiving-end  current  and  lagging  90 


degrees  behind  it  and  the  other  as  the  result  of  the  line 
charging  current  and  lagging  90  degrees  behind  it. 
These  two  combine  at  an  angle,  with  each  other  and 
with  the  impressed  e.m.f.  at  the  sending-end. 

CHARGING  CURRENT 

Conductors  of  a  circuit,  being  separated  by  a  di- 
electric (such  as  air,  in  overhead  circuits,  or  insulation 
in  cables),  form  a  condenser.  When  alternating-cur- 
rent flows  through  such  a  circuit,  current  (known  as 
charging  current)  virtually  passes  from  one  conductor 
through  the  dielectric  to  the  other  conductors,  which 
are  at  a  diflferent  potential.  This  current  is  in  shunt 
with  the  circuit,  and  differs  from  the  current 
which  passes  between  conductors  over  the  insula- 
tors etc.  (leakage  current)  or  through  the  air 
(corona  effect)  only  in  that  the  charging  current 
leads  the  voltage  by  90  degrees,  whereas  the  leak- 
age current  is  in  phase  with  the  voltage. 

For  a  given  spacing  of  conductors,  the  charg- 
ing current  increases  with  the  voltage,  the  fre- 
quency and  the  length  of  the  circuit  For  long 
high-voltage  circuits,  particularly  at  60  cycles 
per  second,  the  charging  current  may  be  as  much 
as  the  full-load  current  of  the  circuit,  or  more. 
In  some  cases  of  long  60  cycle  circuits,  where  a 
comparatively  small  amount  of  power  is  to  be 
transmitted,  it  is  necessary  to  limit  the  voltage  of 
transmission,  in  order  that  the  charging  current 
may  not  be  so  great  as  to  overload  the  genera- 
tors. This  charging  current,  being  in  leading 
quadrature  with  the  voltage,  represents  nearly  all 
reactive  power,  but  it  is  just  as  effective  in  heat- 
ing the  generator  windings  as  if  it  represented 
active  power.  On  the  other  hand,  it  combines 
with  the  receiving-end  current  at  an  angle  (de- 
pending upon  the  power-factor  of  the  receiver 
load)  in  such  a  manner  that  the  addition  of  the 
full-load  receiving-end  current,  in  extreme  cases, 
may  not  greatly  increase  the  sending  end  current. 
In  other  words  (if  the  charging  current  is  near 
full-load  current)  the  current  at  the  generator 
end  may  not  increase  much  when  full  load  at  the 
receiver  end  is  added,  over  what  it  is  when  no 
load  is  taken  off  at  the  receiving-end. 

Since  the  e.m.f.  of  self-induction  due  to  the 
charging  component  is  proportional  to  the  charg- 
ing current,  its  effect  upon  the  voltage  regulation 
of  the  circuit  will  also  be  proportional  to  the  charging 
current.  For  a  short  low-voltage  circuit,  the  charging 
current  is  so  small  that  its  effect  on  voltage  regulation 
may  be  ignored.  On  the  longer  circuits,  especially  long 
60  cycle  circuits,  such  as  will  be  considered  later,  its 
effect  must  be  given  careful  consideration. 

VARIATION   IN    CURRENT  AND  VOLTAGE  ALONG  THE 
CIRCUIT 

It  was  explained  above  and  illustrated  in  Fig.  20 
that  with  a  receiving-end  load  of  leading  power-factor. 


GRAPHICAL   SOLUTION   FOR   LONG   LINES 


63 


the  voltage  at  the  sending-end  of  the  circuit  might  be 
less  than  that  at  the  receiving-end.  It  was  shown  that 
the  e.m.f.  of  self-induction,  resulting  from  the  leading 
current,  tends  to  raise  the  voltage  along  the  circuit. 
This  boosting  effect  of  the  voltage  is  entirely  due  to  the 
leading  component  of  the  load  current. 

If,  now,  it  is  assumed  that  the  power-factor  of  the 
receiving-end  load  is  100  percent,  there  will  be  no  lead- 
ing component  in  the  load  current,  and  therefore  there 
can  be  no  boosting  of  the  voltage  due  to  the  load  cur- 
rent. Since,  however,  all  circuits  have  capacitance, 
and  since  the  current  is  alternating,  charging  current 
Vv'ill  flow  into  the  line  and  this  being  a  leading  current, 
the  same  tendency  to  raise  the  voltage  along  the  circuit 
will  take  place  as  is  illustrated  by  Fig.  20. 

The  upper  part  of  Fig.  21  is  intended  to  give  a 
physical  conception  of  what  takes  place  in  an  alternat- 
ing-current circuit.  As  the  load  current  starts  out  from 
the  sending-end,  and  travels  along  the  conductor,  it 
meets  with  ohmic  resistance.  This  is  represented  by  r 
in  Fig.  21.  It  also  meets  with  reactance  in  quadrature 
to  the  current.  This  is  represented  by  jx  in  the  dia- 
gram. Superimposed  upon  this  load  current  is  a  cur- 
rent flowing  from  one  conductor  to  the  others,  in  phase 
with  the  voltage  at  that  point  and  representing  true 
power.  This  current  is  the  result  of  leakage  over  in- 
sulators and  of  corona  effect  between  the  conductors. 
It  is  represented  by  the  letter  g  in  the  diagrams.  Then 
there  is  the  charging  current  in  leading  quadrature  with 
the  voltage.  This  current  does  not  consume  any  active 
power  except  that  necessary  to  overcome  the  resistance 
to  its  flow. 

In  Fig.  21  the  four  linear  constants  of  the  alternat- 
ing-current circuit,  r  representing  the  resistance,  jx  re- 
presenting the  reactance,  g  representing  the  leakage  and 
b  representing  the  susceptance,  are  shown  as  located,  or 
lumped,  at  six  different  points  along  the  circuit.  This 
i.,  as  they  would  appear  in  an  artificial  circuit  divided 
into  six  units.  In  any  actual  line,  these  four  constants 
are  distributed  quite  evenly  throughout  the  length  of 
the  circuit. 

VOLTAGE  AND  CURRENT  DISTRIBUTION  FOR  PROBLEM  X 

The  effect  of  the  charging  current  flowing  through 
the  inductance  of  the  circuit  gives  rise  to  a  very  inter- 
esting phenomenon.  In  order  to  illustrate  this  effect, 
the  current  and  voltage  distribution  for  a  60  cycle,  1000 
volt,  three-phase  circuit,  300  miles  long,  is  plotted  in 
Fig.  21.  This  circuit  will  be  referred  to  as  problem  X. 
In  such  a  long  60  cycle  circuit,  this  phenomenon  is  quite 
pronounced;  so  that  such  a  problem  serves  well  as  an 
illustration.  The  voltage  and  the  current  have  been 
determined  for  points  50  miles  spart  along  the  circuit. 
Values  for  both  the  current  and  the  voltage  under  zero 
load,  also  under  load  conditions  have  been  plotted.  The 
load  conditions  refer  to  a  receiving-end  load  of  i8ckx) 
kv-a,  at  90  percent  power-factor,  lagging,  60  cycle 
three-phase.  The  voltage  is  assumed  as  being  held 
constant  104000  volts  at  the  receiving-end,  for  both 
?:»'-o  and  full-load  conditions. 


Zero-Load  Conditions — Without  any  load  being 
taken  from  the  circuit,  it  will  be  seen  that  the  charging 
current  at  the  sending-end  approaches  in  value  that 
established  when  under  full  load;  i.e.,  94.75  amperes. 
The  charging  current  drops  down  to  approximately  50 
amperes  at  the  middle,  and  to  zero  at  the  receiving-end 
of  the  unloaded  circuit.  The  lower  full  line  curve 
shows  how  this  current  is  distributed  along  the  circuit 
Starting  at  zero,  at  the  receiving-end  of  the  circuit,  it 
increases  as  the  sending-end  of  the  circuit  is  ap- 
proached, at  which  point  it  reaches  its  maximum  value 
cf  87.89  amperes.  The  voltage  distribution  under  zero- 
load  conditions  is  some-what  opposite  to  that  of  the  cur- 
rent distribution.  That  is  the  voltage  ( 104  000  volts  at 
the  receiving-end)  keeps  falling  lower  until  it  reached  a 
value  of  84  676  at  the  sending-end.  It  should  be  noted 
that  the  voltage  curve  for  zero  load  condition  drops 
down  rapidly  as  the  sending-end  is  approached.  The 
reason  for  this  is  the  large  charging  current  flowing 
through  the  inductance  of  the  circuit  at  this  end  of  the 
circuit.  The  larger  the  charging  current  the  greater 
the  resultant  boosting  of  the  receiving-end  voltage. 

Load  Conditions — When  16  000  kv-a  at  90  percent 
power-factor  lagging  is  taken  from  the  circuit  at  the 
receiving-end,  the  current  at  this  end  goes  up  to  99.92 
amperes.  As  the  supply  end  is  approached  the  current 
becomes  less,  reaching  its  lowest  value  (approximately 
83  amperes)  in  the  middle  of  the  circuit.  At  the  supply 
end  it  is  94.75  amperes,  which  is  less  than  it  is  at  the 
receiver  end.  Thus  the  full  line  representing  the  cur- 
rent in  amperes  along  the  circuit  assumes  the  form  of  an 
arc,  bending  downward  in  the  middle  of  the  circuit. 
The  shape  of  this  current  curve  is  dependent  upon  the 
relative  values  of  the  leading  and  lagging  components 
cf  the  current  at  points  along  the  circuit.  The  reason 
that  the  current  is  a  minimum  rear  the  middle  of  the 
circuit,  is  because  this  is  the  point  where  the  lagging 
current  of  the  load  and  the  leading  charging  current  of 
the  circuit  balance  or  neutralize  each  other,  and  the 
power-factor  is  therefore  unity.  Starting  at  the  re- 
ceiving-end, the  power- factor  is  90  percent  lagging.  As 
the  middle  of  the  circuit  is  approached,  the  increasing 
charging  current  neutralizes  an  increasing  portion  of  the 
lagging  component  of  the  load  current.  Near  the 
middle  of  the  circuit,  this  lagging  component  is  en- 
tirely neutralized,  and  the  power-factor  therefore  rises 
to  unity.  Passing  the  middle  and  approaching  the 
sending-end  there  is  no  more  lagging  component  to  be 
neutralized,  and  the  increasing  charging  current 
causes  a  decreasing  leading  power-factor  which,  when 
the  sending-end  is  reached,  becomes  93.42  percent 
leading.  It  will,  therefore,  be  seen  that  the  power- fac- 
tor as  well  as  the  current  and  voltage  varies  through- 
nut  the  length  of  the  circuit. 

The  voltage  distribution  under  load  condition  is 
indicated  by  the  top  broken  line.  In  order  that  the  re- 
ceiving-end voltage  may  be  maintained  constant  at 
104000  volts,  the  voltage  at  the  sending-end  will  vary 


64 


GRAPHICAL  SOLUTION  FOR  LONG  LINES 


from  84  676  volts  at  zero  load  to  122  370  volts  at  the 
pssumed  load. 

THE    AUXILIARY    CONSTANTS 

With  the  impedance  methods  considered  under  the 
general  heading  of  "Short  Transmission  Lines"  the  cur- 
rent vifas  considered  as  of  the  same  value  throughout 
the  circuit,  and  the  voltage  drop  along  the  circuit  was 
considered  as  proportional  to  the  distance.  These  as- 
sumptions, virhich  are  permissible  in  case  of  short  Imes, 
are  satisfied  by  simple  trigonometric  formulas. 

The  rigorous  solution  for  circuits  of  great  elec- 
trical length  accurately  takes  into  account  the  effect 
produced  by  the  non-uniform  distribution  of  the  cur- 
rent and  the  voltage  throughout  the  length  of  the  circuit. 
This  effect  wfill  hereafter  be  referred  to  as  the  distribu- 
tion effect  of  the  circuit,  and  may  be  taken  into  account 


SENDING 
END     E3 


EQUIVALENT  CIRCUIT  TO  NEUTRAL 
r    INDICATES  RESISTANCE    jx  REACTANCE    jb  SUSCEPTANCE   AND  g  LEAKANCE 

(P.F.  OF  LOAD.90%  LAGGING) 
_VOLTAOE  (LOAD  CONDITION) 


ti! 

1 20s 


SENDINO-END 


100  l&O  '^00 

DISTANCE  IN  MILES  FROM  SENDING-END 


FIG.  21 — DIAGRAMS   OF  TRANSMISSION   CIRCUIT — PROBLEM   X 

300  miles  long,  104000  volts  delivered,  60  cycle.  The  upper  diagram  gives  a 
physical  conception  of  the  conditions  along  the  line.  The  curves  show  the  variation 
in  current  and  voltage  along  the  circuit. 

through  the  application  of  the  so  called  auxiliary  con- 
stants of  the  circuit. 

The  auxiliary  constants  A,  B  and  C  of  the  circuit 
are  functions  of  its  physical  properties,  and  of  the  fre- 
quency only.  They  are  entirely  independent  of  the 
voltage  or  current  of  the  circuit.  The  various  solu- 
tions for  long  transmission  circuits  are  in  effect 
schemes  for  determining  the  values  of  these  three 
auxiliary  constants.  Mathematically  they  may  be  cal- 
culated, by  hyperbolic  functions  or  by  their  equivalent 
convergent  series.  Graphically  they  may  be  obtained 
to  a  high  degree  of  accuracy  from  the  accompanying 
Wilkinson  Charts  for  overhead  circuits  not  exceeding 
300  miles  in  length.  Having  determined  the  values  for 
these  three  constants  for  a  given  circuit,  the  remainder 
of  the  solution  is  just  as  simple  as  for  short  lines  It 
is  only  necessary  to  apply  any  desired  load  conditions 
to  these  constants  and  plot  the  results  by  vector  dia- 
grams. 


DIAGRAM    OF  THE  AUXILIARY   CONSTANTS 

In  Fig.  22  are  shown  voltage  and  current  diagrams 
lepresenting  the  application  of  the  auxiliary  constants 
to  the  solution  of  transmission  circuit  problems.  To 
construct  the  voltage  vector  diagram,  the  two  auxiliary 
constants  A  and  B  are  required,  and  to  construct  the 
current  vector  diagram,  constants  A  and  C  are  required. 
Since  these  diagrams  are  based  upon  one  volt  and 
one  ampere  at  the  receiving-end,  it  is  necessary  to 
multiply  the  values  of  the  auxiliary  constants  by  the 
volts  or  the  amperes  at  the  receiving-end,  in  order  to 
apply  the  auxiliary  constants  to  a  specific  problem. 
Since  the  diagrams  are  shown  corresponding  to  unity 
power-factor,  it  will  also  be  necessary  to  change  the 
position  of  the  impedance  and  charging  current  triangles 
in  case  the  power-factor  differs  from  unity.  This  will 
be  explained  later. 

Constants   a■^   and   a^ — Refer- 
ring to  the  voltage  diagram,  Fig. 
22,  if  the  line  is  electrically  short 
the  charging  current,  and  conse- 
quently its  effect  upon  the  volt- 
age regulation  is  small.     In  such 
a  case  the  auxiliary  constant  o^ 
would  be  unity,  and  the  auxiliary 
constant  aj  would  be  zero.    In 
other  words,  the  impedance  dia- 
gram    would     (for      a  power- 
factor     of     100     percent)     be 
built     upon     the     end     of     the 
vector   ER,    the   point    0    coin- 
ciding with  the  point  R.  In  such 
a  case,  the  voltage  at  the  send- 
ing end,  at  zero  load,  would  be 
the  same  as  that  at  the  receiving- 
end.     If  the  circuit  contains  ap- 
preciable capacitance,  the  e.m.f. 
of  self-induction,  resulting  from 
the  charging  currents  which  will 
flow,    will    result    in    a    lower    voltage    at    zero    load 
at      the     sending-end     than     at     the      receiving-end 
of    the    line,    as    previously    explained.       Obviously, 
the     load     impedance     triangle     must     be     attached 
to  the  end  of  the  vector  representing  the  voltage  at  the 
sending-end  of  the  circuit  at  zero  load.     This  is  the 
vector  EO  of  the  voltage  diagram.  Fig.  22.     This  volt- 
age diagram  corresponds  to  that  of  a  60  cycle  circuit, 
300  miles  in  length.     In  such  a  circuit,  the  effect  of  the 
charging  current  is  sufficiently  great  to  cause  the  shift- 
ing of  the  point  0  from  R  (in  a  short  line)  to  the  posi- 
tion shown  in  Fig.  22.     In  other  words,  the  voltage  at 
zero  load  at  the  sending-end  has  shifted  from  ER  for 
circuits  of  short  electrical  length,  to  EO  for  this  long 
60  cycle  circuit.     The  auxiliary  constants  a^  and  a,, 
therefore,  determine  the  length  and  position  of  the  vec- 
tor representing  the  sending-end  voltage  at  zero  load. 
Actually,  the  constant  a^  represents  the  volts  resistance 
drop  due  to  the  charging  current,  for  each  volt  at  the 


300  VllLES 
RECEIVING-END 


GRAPHICAL  SOLUTION  FOR  LONG  LINES 


65 


receiving-end  of  the  circuit.  That  is,  the  line  OF  equals 
approximately  one-half  the  charging  current  times  the 
resistance  R,  taking  into  account,  of  course,  the  dis- 
tributed nature  of  the  circuit.  If  the  circuit  is  short,  it 
would  be  sufficiently  accurate  to  assume  that  the  total 
charging  current  flows  through  one-half  of  the  resist- 
ance of  the  circuit.  To  make  this  clear,  it  will  be  shown 
later  that,  for  problem  X,  the  resistance  per  conductor 
R  =  105  ohms  and  the  auxiliary  constant  C,  = 
0.001463.  Thus,  this  line  will  take  0.001463  ampere 
charging  current,  at  zero  load,  for  each  volt  maintained 
at  the  receiving-end,  and  since  OF  =  approximately 

R                                                          105 
/c  X    —  \ve  have  OF  (a^)   =  0.001463  X = 

0.0768075.  The  exact  value  of  O2  as  calculated  rigor- 
ously, taking  into  account  the  distributed  nature  of  the 
circuit,  is  0.076831.     Since  the  charging  current  is  in 

s 


VOLTAGE   DIAGRAM 


<- VOLTAGE  AT  RECEIVING-EN0=ONE  VOLT- 

H 


CURRENT  DIAGRAM 


-CURRENT  AT  RECEIVING -END=ONE  AMPERE 


FIG.    22 — DI.\GR.'\MMATIC  REPRESENTATION  OF  AUXILIARY  CONSTANTS 
OF  A  TRANSMISSION  CIRCUIT 

The  vectors  are  based  upon  one  volt  and  one  ampere  being 
delivered  to  the  receiving  end  at  unity  power-factor.  These 
diagrams  correspond  to  those  of  a  long  circuit, 
leading  quadrature  with  the  voltage  ER,  the  resistance 
drop  OF  due  to  the  charging  current  is  also  at  right 
angles  to  ER,  as  in  Fig.  22. 

The  length  of  the  line  FR  or  (/ — Oj),  represents 
the  voltage  consumed  by  the  charging  current  flowing 
through  the  indi^ctance  of  the  circuit.  This  may  also 
be  expressed  with  small  error  if  the  circuit  is  not  of 

X 
great  electrical  length  as  /o  X  — •     The  reactance  per 

conductor  for  problem  X  is  249  ohms.  Therefore  FR  = 
249 


0.001463   X 


0.182143  and  Oi  =:   i.oooooo 


0.182143  =  0.817857.  The  exact  value  for  o,  as  cal- 
culated rigorously,  taking  into  account  the  distributed 
nature  of  the  circuit,  is  0.810558.  The  vector  FR,  re- 
presenting the  voltage  consumed  by  the  charging  cur- 
rent flowing  through  the  inductance,  is  naturally  in 
quadrature  with  the  vector  OF,  representing  the  voltage 
consumed  by  the  charging  current  flowing  through  the 
resistance  of  the  circuit. 


Constants  b^  and  b^  represent  respectively  the  re- 
sistance and  the  reactance  in  ohms,  as  modified  by  the 
distributed  nature  of  the  circuit.  The  values  for  these 
constants,  multiplied  by  the  current  in  amperes  at  the 
receiver-end  of  the  circuit,  give  the  IR  and  IX  volts 

CONSTANT  (A) 


~60       Too        BO      200      555      5o5      350     400      460      600 
TRANSMISSION  DISTANCE— MILES 


CONSTANT  (B) 


25  CYCLtS  bj 
J5  CVCLES^ 

SO  CYCLES  b} 


SO  CYCLES  b| 


100       1 50      200      250      300      360      400      460     600 
TRANSMISSION  DISTANCE-MILES 


CONSTANT  (C) 


100        150       MO       250       300       360      400 
TRANSMISSION  DISTANCE-MILES 

FIG.   23 — VARIATION    OF   THE    AUXILIARY    CONSTANTS    FOR   CIRCUITS 
OF  DIFFERENT  LENGTHS 

drop  consumed  respectively  by  the  resistance  and  the 
reactance  of  the  circuit.  To  illustrate  this,  the  values 
of  R  and  X  for  problem  X  are  i?  =  105  ohms  and  X  = 
249  ohms  per  conductor.  The  distribution  effect  of  the 
circuit  modifies  these  linear  values  of  R  and  A'  so  that 


66 


GRAPHICAL  SOLUTION  FOR  LONG  LINES 


their  effective  values  are  b^  =  91.7486  and  b^  = 
235.868  ohms.  The  impedance  triangle,  as  modified  so 
as  to  take  into  exact  account  the  distributed  nature  of 
the  circuit,  is  therefore  smaller  than  it  would  be  if  the 
circuit  were  without  capacitance. 

Constants  c^  and  c^  represent  respectively  conduct- 
ance and  susceptance  in  mhos  as  modified  by  the  dis- 
tributed nature  of  the  circuit.  The  values  for  these 
constants,  multiplied  by  the  volts  at  the  receiving-end 
of  the  circuit,  give  the  current  consumed  respectively 
by  the  conductance  and  the  susceptance  of  the  circuit. 
To  illustrate,  the  value  of  B  for  problem  X  is  0.001563 
mho  per  conductor.  The  distribution  effect  of  the  cir- 
cuit modifies  this  fundamental  value  so  that  its  effec- 

CONSTANT  (A)=(a|i-ja;) 


V4TH  wave       1/2  WAVE        %TH  WAVE    FULL  WAVE 


FIG.    24 — VARIATION   OF   THE   AUXILIARY    CONSTANTS 

For  a  60  cycle  circuit  (problem  X)  up  to  full  wave  length, 
tive  value  c^  =  0.001463.  The  value  of  Cj  is  so  small 
that  its  effect  is  negligible  for  all  except  very  long  cir- 
cuits. For  power  circuits  it  will  usually  be  sufficiently 
accurate  to  neglect  c^.  The  value  c^  will  in  such  cases 
represent  the  charging  current  at  zero  load  per  volt  at 
the  receiving-end.  Thus  c^,  multiplied  by  the  receiving- 
end  voltage,  gives  the  charging  current  at  zero  load  for 
the  circuit.  For  problem  X,  c^  =  0.001463,  and  this, 
multiplied  by  the  receiving-end  voltage  to  neutral 
60044  =  87.85  amperes  charging  current  per  con- 
ductor. 

VARIATION    IN   THE   AUXILIARY    CONSTANTS 

The  curves.  Fig.  23,  will  serve  to  illustrate  in  a 
general  way  how  the  auxiliary  constants  vary  for  t^oth 


25  and  60  cycle  circuits  for  lengths  up  to  and  includ- 
ing 500  miles.  In  other  words  these  curves  have  been 
plotted  from  calculated  values  for  these  constants  for 
certain  circuits. 

When  the  circuit  is  short,  these  constants  do  not 
vary  materially  from  the  linear  constants  of  the  cir- 
cuit, but  when  the  circuit  becomes  long,  they  depart 
rapidly,  particularly  if  the  frequency  is  high. 


AUXILIARY 
OONSTANTS 

WAVE  LENGTH  OF  THE  CIRCUIT  AND  TRANSMISSION  DISTANCE-MILES         [ 

'/sTH 

'A 

.  %TH 

Vt 

.VaTH 

% 

'/sTH, 

FULL 

369.9 
MILES 

1739.8 

.MILES' 

III09.7 
MILES 

1479.5 
MILES. 

1849.4 
MILES, 

2219.3 
.MILES 

2589.2 
MILES 

2959   1 
MILES 

■a, 

82 

f  .7/4 
+  ,//3 

0 
+.323 

-.789 
+  .350 

0 

-■9-f2 
-.622 

0 
-/.  /O-f 

+  /./9/ 
-.938 

+1.922 
0 

*  105 

••26; 

*  87 

+  428 

-77.5 
*3S0 

-27* 
+  SS.S 

-330 
-330 

-/22 
-60s 

+  293 
-S60 

+  &70 
-/3S 

0, 

-.000075 
+.00/7^ 

-.00060 

+.OOX.'*? 

-oo/z 
+.00169 

'OPth 
-.0003Z2 

-,00101 
-.00Z50 

■*-.ooo7/ 
-.003S 

+.O028 
-.00233 

+.0039 

•^.ooo78 

(A) 

.72  5 

.323 

/9o°oo' 

.843 
//Si'  OS' 

I.Z09 

//So'oo' 

I.IZ9 

/zi-fzC 

/./04 
/27d'00' 

'■SZ8 
/32l°ll' 

I.9ZZ, 
/3i<fOO 

(B) 

30I--* 

.+  37 

3se.s 

//02°2?' 

ZB2.3 
//(S°3l' 

■H9.S 
/ZZS°07 

*/9.^ 
/S5S°3V 

^35.7 
/2  9  7=23 

iSZ.t 
/345:3.»' 

(C) 

.00'743 

•002527 
//o/°2«' 

-OOZO7S 
/IZi°ZI- 

.00/633 
//9/'Z(; 

.0027/i" 
/Z17'39' 

.003J82 
/zil'Zi 

.O03677 
/330'/S' 

.003947 
/37/''2«' 

FIG.    25 — VAHIATION   OF   THE  AUXILIARY   CONSTANTS 

For  problem  X  up  to  full  -wave  length. 

The  auxiliary  constants  have  been  calculated  for 
problem  A'  up  to  and  including  a  full  wave  length, 
namely  2959  miles.  Calculations  were  made  only  for 
distances  representing  each  i/8xh.  wave,  that  is  each 
370  miles.  The  results  are  tabulated  in  Fig.  25,  and 
are  plotted  graphically  in  Fig.  24.  It  is  interesting  to 
note  how  these  auxiliary  constants  vary  with  increasing 
negative  and  positive  values  as  the  circuit  increases  in 
length.  A  polar  diagram  is  plotted  in  Fig.  26,  indicat- 
ing the  manner  in  which  the  auxiliary  constant  A  and 
its  rectangular  co-ordinates  vary.  Although  these  ex- 
treme variations  are  instructive  and  interesting,  ^hey 
are  not  encountered  in  power  transmission  circuit:;,  al- 
though they  will  be  in  long  distance  telephone  practice. 


Vi  WAVE  LENGTH, 
739.8  MILES 


%TH  WAVE  LENGTH 
1109.7  MILES, 


V2  WAVE  LENGTH 
1479.6  MILES 


'/4th  wave  LENGTH 

369.9  MILES 

i>  ,FULL  WAVE  LENGTH 
2959.1  MILES 


T2.0 
CONSTANT  (a) 


%TH  WAVE  LENGTH 
I  849.4  MILES 


%TH  WAVE  LENGTH 
2589.2  MILES 


3/4  WAVE  LENGTH 
2219.3  MILES 


•  •M .  6- 
FIG.    26 — POLAR  DIAGRAM 

Showing    the    variation    of    the   auxiliary   constant   A    for 

hlf*m      y       nr»     *-n      full     timro     1at->n-fV. 


.^..lu  ,.  »»'fe  Vil^,  Ytli  lCH.i»_»ll         Ul  lilC 

problem  X,  up  to  full  wave  length. 


THE    WILKINSON    CHARTS 

Mr.  T.  A.  Wilkinson  has  prepared  charts  frum 
v.hich  the  auxiliary  constants  may  be  read  directly,  thus 
abridging  a  great  amount  of  tedious  mathematical  cal- 
culation. These  charts,  are  plotted  for  circuits  of 
lengths  up  to  and  including  300  miles.* 


♦Similar  Charts  by  Mr.   Wilkinson  were  published  in  the 
Electrical  World  for  Mar.  16,  igi8. 


GRAPHICAL  SOLUTION  FOR  LONG  LINES 


67 


CHART  V— WILKINSON  CHART  A 

(FOR  DETERMININQ  AUXILIARY  CONSTANTS-ZERO  LOAD  VOLTAGE) 
o 


00  COOT  en  O  (X  O 

CVCLES  RESISTANCE  (OHMS)  X  SUSCEPTANCE    ( MlCROr."  lOS)-  PER  MIUE 


66 


GRAPHICAL  SOLUTION  FOR  LONG  LINES 


their  effective  values  are  b^  =  91.7486  and  b^  = 
235.868  ohms.  The  impedance  triangle,  as  modified  so 
as  to  take  into  exact  account  the  distributed  nature  of 
the  circuit,  is  therefore  smaller  than  it  would  be  if  the 
circuit  were  without  capacitance. 

Constants  c^  and  c^  represent  respectively  conduct- 
ance and  susceptance  in  mhos  as  modified  by  the  dis- 
tributed nature  of  the  circuit.  The  values  for  these 
constants,  multiplied  by  the  volts  at  the  receiving-end 
of  the  circuit,  give  the  current  consumed  respectively 
by  the  conductance  and  the  susceptance  of  the  circuit. 
To  illustrate,  the  value  of  B  for  problem  X  is  0.001563 
mho  per  conductor.  The  distribution  effect  of  the  cir- 
cuit modifies  this  fundamental  value  so  that  its  effec- 


COrslSTANT  (A)  =  (a|  <-ia2) 


'/4THWAVE       1/2  WAVE        %THWAVE    FULL  WAVE 


KIG.   24 — VARIATION   OF   THE   AUXILIARY   CONSTANTS 

For  a  60  cycle  circuit  (problem  X)  up  to  full  wave  length, 
tive  value  c^  =  0.001463.  The  value  of  c^  is  so  small 
that  its  effect  is  negligible  for  all  except  very  long  cir- 
cuits. For  power  circuits  it  will  usually  be  sufficiently 
accurate  to  neglect  c^.  The  value  c^  will  in  such  cases 
represent  the  charging  current  at  zero  load  per  volt  at 
the  receiving-end.  Thus  c^,  multiplied  by  the  receiving- 
end  voltage,  gives  the  charging  current  at  zero  load  for 
the  circuit.  For  problem  X,  c^  =  0.001463,  and  this, 
multiplied  by  the  receiving-end  voltage  to  neutral 
60044  ^=  87.85  amperes  charging  current  per  con- 
ductor. 

VARIATION   IN   THE   AUXILIARY    CONSTANTS 

The  curves.  Fig.  23,  will  serve  to  illustrate  in  a 
general  way  how  the  auxiliary  constants  vary  for  t^oth 


25  and  60  cycle  circuits  for  lengths  up  to  and  includ- 
ing 500  miles.  In  other  words  these  curves  have  been 
plotted  from  calculated  values  for  these  constants  for 
certain  circuits. 

When  the  circuit  is  short,  these  constants  do  not 
vary  materially  from  the  linear  constants  of  the  cir- 
cuit, but  when  the  circuit  becomes  long,  they  depart 
rapidly,  particularly  if  the  frequency  is  high. 


AUXILIARY 
OONSTANTS 

WAVE  LENGTH  OF  THE  CmOUIT   AND  TRANSMISSION  DISTANCE-MILES          | 

'/sTH 

'A 

.  %TH 

'/2 

.%TH 

% 

'/sTH, 

FULL 

369.9 

MILES 

1739  8 
MILES' 

III09.7 
MILES 

1479.6 
MILES, 

1849.4 
MILES, 

2219  3 
.MILES 

2589.2 

MILES 

2959.1 
MILES 

■ai 

+  .7/i 

0 
+.323 

-.789 
+  .350 

-/.  20? 
0 

-■942 
-.iZI 

0 
-I.I  Of 

*-l.l9l 
-.953 

+  /.922 
0 

&.■ 

*■  lOS 

<-28; 

+  87 
+  428 

-77.5 
+  350 

-Z7i 

*ss.s 

-330 
-330 

-IZi. 
-60s 

+  Z92 
-SiO 

+  &70 
-I3S 

0, 

-.000075 
+.00/7'* 

-.00050 

-oo/a 

-onik 
-.000322 

-.OOIOI 

-.00250 

+.0007i 

-.0035 

+.O028    *:0039 
-.00233 +.0007S 

(A) 

.7ZS 

.3Z3 
/  90^00 

.8^3 

lis(,°  oi 

1.2.09 

//■ao'oo 

I.IZ9 
/2/3°24' 

1.104 

/Z7O°C0' 

'■SZS 
/32l°ll' 

I.92Z, 
/3t0°OO 

(B) 

301.* 

1-37 
/79°3^' 

358.S 
//02°2'7' 

282.3 

/ZUg'oT 

il9.3 
/sS8«3-f 

i-iS.7, 

/z^rz-i 

i^2.4 
/348-3-f' 

(C) 

■  oon-t-i 

■002527.  Ooa07i' 

.00/433 
//?/°26' 

.0027/5 

/217'rf 

.0O3582 
/2^l'2f 

.003i77 
/320'IS' 

■O03947 
/37l°2i' 

FIG.   25 — VARIATION   OF   THE   AUXILIARY   CONSTANTS 

For  problem  X,  up  to  full  wave  length. 

The  auxiliary  constants  have  been  calculated  for 
problem  X  up  to  and  including  a  full  wave  length, 
namely  2959  miles.  Calculations  were  made  only  for 
distances  representing  each  i/d'th  wave,  that  is  each 
370  miles.  The  results  are  tabulated  in  Fig.  25,  and 
are  plotted  graphically  in  Fig.  24.  It  is  interesting  to 
note  how  these  auxiliary  constants  vary  with  increasing 
negative  and  positive  values  as  the  circuit  increases  in 
length.  A  polar  diagram  is  plotted  in  Fig.  26,  indicat- 
ing the  manner  in  which  the  auxiliary  constant  A  and 
its  rectangular  co-ordinates  vary.  Although  these  ex- 
treme variations  are  instructive  and  interesting,  Ihey 
are  not  encountered  in  power  transmission  circuits,  al- 
though they  will  be  in  long  distance  telephone  practice. 


'/4  WAVE  LENGTH. 
739.8  MILES 


%TH  WAVE  LENGTH 
MOB. 7  MILES. 


Vl  WAVE  LENGTH 
1479  5  MILES 


%TH  WAVE  LENGTH 
I  849.4  MILES 


■  •/gTH  WAVE  LENGTH 
369.9  MILES 

P  /FULL  WAVE  LENGTH 
2969.1  MILES 


+2.0 
CONSTANT  (a) 


%TH  WAVE  LENGTH 
2589.2  MILES 


3^  WAVE  LENGTH 
2219.3  MILES 


FIG.   26 — POLAR  DIAGRAM 

Showing  the  variation  of  the  auxiliary  constant  A  for 
problem  A',  up  to  full  wave  length. 

THE    WILKINSON    CHARTS 

Mr.  T.  A.  Wilkinson  has  prepared  charts  frum 
v.hich  the  auxiliary  constants  may  be  read  directly,  thus 
abridging  a  great  amount  of  tedious  mathematical  cal- 
culation. These  charts,  are  plotted  for  circuits  of 
lengths  up  to  and  including  300  miles.* 


♦Similar  Charts  by  Mr.   Wilkinson  were  published  in  the 
Electrical  World  for  Mar.  16,  1918. 


GRAPHICAL  SOLUTION  FOR  LONG  LINES 


fi7 


CHART  V— WILKINSON  CHART  A 

(FOR  DETERMININQ  AUXILIARY  CONSTANTS-ZERO  LOAD  VOLTAGE> 


CVCLES  RESIST ANCe""(0HMS)  X  SUSCtPTANCE    (MICR0K;M0S)-PER  MILE 


68  GRAPHICAL  SOLUTION  FOR  LONG  LINES 

CHART    VI— WILKINSON    CHART    B 

0.90      I  op 


(FOR  DETERMINING  AUXILIARY  CONSTANTS-LINE  IMPEDANCE) 
2601 


CHART   B 

PROBLEM 
FIND  THE  AUXILIARY  CONSTANTS  FOR  A  TH^EE  PHASE  80  CVCLE 
TRANSMISSION  CIRCUIT  300  MILES  LONO    GON3I8TINO  OF    THSEE 
*000  STRANDED  COPPER  CONDUCTORS  WITH  lO'  X  Kl'  X  20'  FLAT  SPACING 
-  ^10X10X20-13.6-  EQUIVALENT  DELTA  SPACINGl-FROM  WIRE 

TABLES  THE  FOLLOWING  VALUES  PER  MILE  Of  S'NGlE  CONDUCTOR  ARE 
OBTAINED,    r-0.3B  OHM.  X- 0.83  OHM 

TO  FIND  THE  CONSTANTS  FROM  THE  CHART. 
CONSTANT "b"- FROM  THE  DOT  AT  THE  RIGHT  AT  THE  INTERSECTION  C 
LINES  FOfl  300  MILES  AND  60  CYCLES  FOLLOW  HORIZONTALLY  TO  THE 
RiCHT  TO  AN  INTERSECTION  WITH  THE  DIAGONAL  FOR  F-  0  35 
VERTICALLY  ABOVE  THE  LATTER  POINT  READ  THE  VALUE  OF  "b"-  D  I  7 
CONSTANT  "b^  -FROM  THE  DOT  ABOVE  AT  THE  INTERSECTION  OF  LINES 
FOR  300  MILES  AND  60  CYCLES  FOLLOW  VERTICALLY  UPWARD  TO  AN 
INTERSECTION  WITH  THE  DIAGONAL  FOR  X-0  83    HORIZONTALLY  TO 
THE  RIGHT  OF  THE  LATTER  PQINT  READ  THE  VALUE  OF  "b'l  ■■  236. 
CONSTANT  (B)-  THIS  IS  THE  TQTAl  IMPEDANCE  PER  CONDUCTOR  TO 
THE  LOAD  CURRENT     IT  MAY  BE  READ  FROM  THE  CHART  AS  FOLLOWS 
ON  THE  CIRCUUR  ARC  IN  THE  UPPER  RIGHT  HAND  QUADRANT  AT  THE 
INTERSECTION  OFCOORDINATEa"bj-Bl  7  ANo"b'j- 336  HEAD  THE 
VALUE  OF  iai->2b2. 

OOPVRIOHT  I9ia  BY  T,  K  WILKINSON 


CYCLES 


0        0.15         0.20         0.25         0.30 
RESISTANCE  (OHMS)-PER  MILE 


GRAPHICAL  SOLUTION  FOR  LONG  LINES 


69 


CHART  VII— WILKINSON   CHART  C 

(FOR  DETERMINING  AUXILIARY  CONSTANTS-CHARGING  CURRENT) 


0019- 


BOO 


CHART  C 

wtoeiEM 

riNO  TMt  4UKH.IW*  CONSTANTS  FQI  *  TMftEE  PMASC  CO  CClC 
TR*f*SMiSSiON  C'«CU'T  300  Mn.es  LOhC  consisting  Of  TMBCC 
♦000  8TBANDC0  COPPtC  CON0UCTO«S  WiTm  pO'X  iO«  W'LAT  s^fcC-MJ 

1-       .jToTrSTjo  -  I  3  8  couivw.cntoclt»sp*c«»«<-»»om-«c 

TABLES  THE  FCXLQWtNO  VALUES  PC"  M'LE  OT  SINGLE  CONOUCTO«  W>E 
OBTAINED     r-0.36OHiyl.b-»  2'"'C*'0*'**0S^0'«"'"*»-     THe«tfO«C 

rb'.oasKs  7>^  b  w 

TO  f  INO  THE  CONSTANTS  raQM  T>^  CWMT 
qpNST ANT "c"- f  WOM  THE  OCT  AT  TmC  tCFT  AT  T*«C  »iTEItS£CTK>«Of 
LINE  S  FOft  JOO  MILES  AND  80  CtClES  f  OLLOw  mOBi20NTALL'  TO  '►<  lE'T 
TO»N  iNTERSeCTONVWlTH  Th|  CHACON *L  fO*»  Ht   -  »  *0     Vt»»TlC*H.* 
ABOVE  TmE  LATTEB  P0<NT  ftCAO  TmE  VALUE  O'  "C,  .  -  0  0000« 
gQfsiSTANT  -C,  -tnOU  TNt  DOT  ABOvt  AT  TmC  inTERSCCTONOT  imtS 
roe  JOO  MILES  *N0  M  CYCLE S  'OLLOW  wCftTtCALLV  U»^W*«0  TO  *N  INTER- 
SECTION *nTM  The  DIAGONAL  FQA  b-  S  3 1     mO«U'ONTA4.L»  TO  ThC  tC'  T  0* 
THE  LATTER  POINT  READ  The  VALUE  Of  "Cj-OOO'M 
f^ONSTAtsiT  (C)  -  THtS  IS  THE  TOTAL  ChARCnC  CuMCNT  RCR  CONOUCO" 
»»£RvOLr  TONEuTRAl  AT  The  RECEIVER  END     WiTntN  PRACTCAL  L«*«TS  ThJ 
VALUE  Of  'C"  WILL  NEVER  t  «CECO  s  "^  VAluC  Of  "Cy     TmC  Cff  ECT  C 
"CI'ON  TMC  NUMERICAL  VALUE  Of  ' Ci  iS  TMERtf ORE  NECl-CBlE  AND 'C' M*» 

BE   TAKEN  AS  EQUAL   TO"Cj  -0  00<*«, 

C0»tr>CmT  )••»  B»  r  A  i*K  "  "NSON 


70 


GRAPHICAL  SOLUTION   FOR   LONG  LINES 


The  reading  of  these  charts  is  simplified  by  reason 
of  the  fact  that  all  three  charts  are  somewhat  similar. 
In  following  any  of  therii,  the  start  is  made  from  the 
intersection  of  the  short  arc  representing  length  of  cir- 
cuit and  the  straight  line  representing  the  frequency. 
From  this  intersection  a  straight  line  is  followed  tc  a 
diagonal  line  and  thence  at  right  angles  to  the  constant 
required.  Thus  in  a  few  minutes  the  auxiliary  con- 
stants of  the  circuit  may  be  obtained  directly  from  the 
chart,  whereas  by  a  mathematical  solution  from  15 
minutes  to  an  hour  might  be  consumed  in  obtaining 
them.  It  is  not,  however,  the  time  saved  in  obtaining 
these  constants  which  is  most  important.  The  greatest 
advantage  in  this  graphical  solution  for  the  auxiliary 
constants  is  that  it  not  only  abridges  the  use  of  a  form 
of  mathematics  which  the  average  engineer  is  inefficient 
in  using,  but  it  tends  to  prevent  serious  mistakes  being 
made.  In  calculating  these  auxiliary  constants  by 
either  convergent  series  or  hyperbolic  methods,  an  in- 
correct algebraic  sign  assigned  to  a  number  may  cause 
a  very  serious  error.  Errors  of  magnitude  are  less 
likely  to  occur  when  using  a  comparatively  simple 
graphical  solution. 

In  order  to  determine  the  accuracy  obtainable  by  a 
complete  graphical  solution,  using  the  Wilkinson  Charts 
for  obtaining  the  auxiliary  constants  and  vector  dia- 
grams for  the  remainder  of  the  solutions,  48  problems 
were  solved  both  graphically  and  mathematically. 
These  problems  consisted  of  circuits  varying  between 
20  and  300  miles  in  length,  and  voltages  varying  be- 
tween 10  000  and  200000  volts.  Twenty- four  prob- 
lems were  for  25  cycle,  and  the  same  number  for  60 
cycle  circuits.  The  maximum  error  in  supply  end  volt- 
='ge  by  the  graphical  solution  employing  a  four  times 
magnifying  glass  was  one-fourth  of  one  percent.  A 
tabulation  of  the  results  as  determined  by  various 
methods  for  these  circuits  will  follow  later. 

APPLICATION  OF  TABLES 

The  application  of  the  tables  to  long  transmission 
lines  follows,  in  general,  the  same  plan  as  for  short 
lines,  published  as  Chart  II,  with  such  modifications  as 
are  produced  by  the  effects  of  distributed  capacitance 
and  reactance.  The  procedure  best  suited  for  long 
transmission  lines  is  shown  in  Chart  VIII. 

GRAPHICAL  SOLUTION  OF  PROBLEM  X 

Problem  X — Length  of  circuit  300  miles,  conduc- 
tors three  No.  000  stranded  copper  spaced  10  by  10  by 
20  feet  (equivalent  delta  12.6  feet)  Temperature  taken 
as  25  degrees  C.  Load  conditions  at  receiving-end 
18000  kv-a,  (16200  kw  at  90  percent  power- factor  lag- 
ging) 104  000  volts,  three-phase,  60  cycles. 

104000 
£r.  = =  60046  volts. 


I, 


1732 
6000  X  1000 


CHART    VIM.— APPLICATION    OF    TABLES    TO 
LONG   TRANSMISSION    LINES 

(EFFECT  OF  DISTRIBUTED   CAPACITANCE  TAKEN  INTO 
ACCOUNT)      OVERHEAD  BARE  CONDUCTORS 


Starting  with   the  kv-a.,   voltage  and  power-factor  at 
the  receiving  end  known. 


QUICK  ESTIMATING  TABLES  XII  TO  XXI  INC. 

From  the  quick  estimating  table  corresponding  to  the 
voltage  to  be  delivered,  determine  the  size  of  the  con- 
ductors corresponding  to  the  permissible  transmission  loss. 


CORONA  LIMITATION— TABLE  XXII 

If  the  transmission  is  at  30000  volts,  or  higher,  this 
table  should  be  consulted  to  avoid  the  employment  of  con- 
ductors having  diameters  so  small  as  to  result  in  excessive 
corona  loss. 


RESISTANCE— TABLE  II 

From  this  table  obtain  the  resistance  per  unit  length 
of  single  conductor  corresponding  to  the  maximum  operat- 
ing temperature — calculate  the  total  resistance  for  one  con- 
ductor of  the  circuit — if  the  conductor  is  large  (250000 
circ.  mils  or  more)  the  increase  in  resistance  due  to  skin 
effect  should  be  added. 


REACTANCE— TABLES  IV  AND  V 

From  one  of  these  tables  obtain  the  reactance  per  unit 
length  of  single  conductor.  Calculate  the  total  reactance 
for  one  conductor  of  the  circuit.  If  the  reactance  is  ex- 
cessive (20  to  30  percent  reactance  volts  will  in  many  cases 
be  considered  excessive)  consult  Table  VI  or  VII.  Hav- 
ing decided  upon  the  maximum  permissible  reactance  the 
corresponding  resistance  may  be  found  by  dividing  this 
reactance  by  the  ratio  value  in  Table  VI  or  VII.  When 
the  reactance  is  excessive,  it  may  be  reduced  by  installing 
two  or  more  circuits  and  connecting  them  in  parallel,  or 
by  the  employment  of  three  conductor  cables.  Using 
larger  conductors  will  not  materially  reduce  the  reactance. 
The  substitution  of  a  higher  transmission  voltage,  with  its 
correspondingly  less  current,  will  also  result  in  less  react- 
ance. 


I 


CAPACITANCE  SUSCEPTANCE- 
IX  AND  X 


-TABLES 


From  one  of  these  tables  obtain  the  capacitance  sus- 
ccptance  to  neutral,  per  unit  length  of  single  conductor. 
Calculate  the  total  susceptance  for  one  conductor  of  the 
circuit  to  neutral. 


GRAPHICAL  SOLUTION 

From  the  Wilkinson  charts  obtain  the  auxiliary  con- 
stants. Applying  these  auxiliary  constants  to  the  load  con- 
ditions of  the  problems,  make  a  complete  graphical  solution 
as  explained  in  the  text.  Vector  diagrams  of  the  voltage 
and  the  current  at  both  ends  of  the  circuit  are  then  con- 
structed, from  which  the  complete  performance  can  be 
readily  obtained  graphically. 


60046 


=  99.92  amperes. 


MATHEMATICAL  SOLUTION 

As  a  precaution  against  errors  in  those  cases  where 
accuracy  is  essential,  the  result  obtained  graphically  should 
be  checked  by  the  convergent  series  or  the  hyperbolic 
method. 


GRAPHICAL   SOLUTION  FOR   LONG   LINES 


71 


From  tables  the  following  linear  constants  per  mile 

are  determined. — 

»"  =  0-35  ohm  (Table  No.  II) 

X  =  0.83  ohm  {Table  No.  V  by  interpolation) 

b  =  521  micromhos  (  Table  No.  X  by  interpolation) 

'  =  {in  this  case  taken  as  zero) 


therefore, 

rb 
and, 

rb'-  — 


=  0-35  X  S-2I  =  1.82 


0.35  X  5-2r  —  950 
The  auxiliary  constants  of  the  above  circuits  are 
now  taken  directly  from  the  Wilkinson  Charts.     This 
problem  is  stated  on  the  Wilkinson  chart.     Following 


AUXILIARY  CONSTANTS  OF  CIRCUIT 

{CALCUUVTEO  RIGOROUSLY  BY  CONVERGENT  SERIES)  " 
(A)-    +  OS  1 006E  ■>  j  007883 1  (B)"=    ♦      91.7488  t  j  238  8880 

(ai  +  jaj)  =  (bi  +  jbj) 

—  8142  /  6"  24'  53'  -263  083    /e8'44' 41' OHM81 


VOLTAGE   DIAGRAM 


(C)  = 


-  CL00004 1 

(o,. 


j(u>oi4e3 

iC2) 


-      0001464  \  91' 36' 36'  MHO 


RECEIVING-END  LOAD   OF     99.92  AMPERES 
AT  90%  P.F    LAGGING 

% 


-48.871  VOLTS  =Ern  XSi 
80.048  VOLTS' 


CURRENT   DIAGRAM 


VECTOR.  OF.  reference; 


S'*-POS1TION  OF  s 
^  FOR  LOAD  PFr 
',\  OF  80*  leading 


FOR  ABOVE 


PROBLEM 
H 


.^-<^ 
.« 


-(O    < 


u1^ 


rent  calculated  rigorously  which  will  appear  in  a  later 
section.  The  /  terms  preceding  some  of  the  numerical 
values  in  Fig.  27  apply  to  the  mathematical  treatment, 
and  have  no  significance  in  connection  with  the 
graphical  solution. 

VOLTAGE  DIAGRAM 

The  vector  ER,  representing  the  constant  voltage 
at  the  receiving-end  (for  all  loads)  is  first  laid  off  to 
some  convenient  scale.  Along  this  vector,  starting  from 
E,  lay  off  a  distance  equal  to  the  receiving-end  voltage 
multiplied  by  the  constant  a^ 
(60046  X  0.810558  =  48,671 
volts).  This  is  EE  of  Fig.  27. 
From  F  lay  off  vertically  (to  the 
same  scale)  the  line  FO  equal  to 
the  receiving-end  voltage  multiplied 
by  the  constant  a^  (60046  X 
0.076831  =  4613  volts).  Connect 
the  points  O  and  £  by  a  line. 
This  line  EO  represents  the  voltage 
at  the  sending-end  at  zero  load. 
This  voltage  vector  may,  if  desired, 
+18.626  VOLTS  bc  locatcd  by  polar  co-ordinates  in 
place  of  rectangular  co-ordinates. 
If  it  is  desired  to  work  with  polar 
co-ordinates  lay  off  the  line  EO  at 
an  angle  of  5°  25'  in  the  forward 
direction  from  the  receiving-end 
voltage  vector  ER.  (For  the  graphi- 
cal solution  it  is  not  necessary  to 
take  account  of  seconds  in  angles) 
The  length  of  the  vector  EO  will 
be  found  by  multiplying  the  con- 
stant A  by  the  receiving-end  volt- 
age   (0.8142   X  60044  =  48889 


^CORRESPONDING 
TOLOAOpr„90% 
LAGGING 


A 


too 
^°o  volts). 


o» 


VECTOR  OF  REFERENCE 


Fir..   27 GRAPHIC   SOLUTION    OF   PROBLEM    X 


the  directions  printed  on  the  charts,  we  obtain  for  this 

circuit  the  following  values  for  the  auxiliary  constants. 

o,  =  0.81      bx  =  91,7    fi  =  0.00004 
02  =  0.077    bs  =^  235     fs  =  0.00146 

From  this  point  on,  the  solution  is  made  graphically 

as  indicated  in  Fig.  27.     It  should  be  noted  here  that 

the   auxiliary   constants   obtained   from   the   Wilkinson 

Charts  are  practically  the  same  as  those  stated  at  the 

top  of  Fig.  27,  which  values  were  calculated  rigorously 

by   convergent   series.     We   will   employ   the   rigorous 

values  in  plotting  the  diagram  so  that  the  values  on  the 

diagram  will  agree  with  the  values  of  voltage  and  cur- 


Having  located  the  point  0,  the 
impedance  triangle  is  built  upon  it 
in  the  following  manner.  Since 
the  power-factor  of  the  load  is  90 
percent  lagging,  determine  from  a 
table  of  cosines  what  the  angle  is 
whose  cosine  is  0.9.  This  is  found 
(from  Table  K)  to  be  25  degrees, 
50  minutes.  Lay  off  the  line  OD 
an  angle  with  the  vector 
25°  50'  in  the  lagging  di- 
of    OD    will    be    determined 


at 
of  reference  ER  of 
rection.  The  length 
by  multiplying  the  current  in  amperes  per  conductor  by 
the  auxiliary  constant  fc,  (9992  X  91.7486  =  9167 
volts).  This  represents  the  resistance  drop  per  con- 
ductor. From  the  point  D  thus  found  draw  a  line  DS 
.it  right  angles  with  OD.  This  line  DS  represents  the 
reactance  volts  per  conductor;  its  length  is  found  by 
multiplying  the  current  in  amperes  per  conductor  by  the 
auxiliary  constant  b^  (99-92  X  235.868  =  23  568  volts). 
Connect  the  point  5  w'ith  £,  the  length  of  which  repre- 
sents the  voltage  (70652  volts)  at  the  sending-end  for 


74 


GRAPHICAL   SOLUTION  FOR  LONG   LINES 

CHART  IX-PETER'S   EFFICIENCY  CHART 

FOR  DETERMINING  TRANSFORMER  LOSSES  AND  EFFICIENCIES 


>i 


EFFICIENCY    IN    PERCENT    AT    LOADS    STATED 

3  I  1 


I: 


FULL 


< 
O 


3 


.i7o 

.2 
.3 
.4 
-.5 
.6 
.7 
.8 
.9 
-1.0 


-1.6- 


-2 


»- 
Z 

UJ 

O 

cc 

UJ 
Q. 


CO 
CO 

o 

_1 
cc 

UJ 

a. 
a. 
o 
o 


-2.5 


-3.5 


-4 


-4.5 


5% 


-99 


■98 


-97 


-96 


-95 


-94 


-93 


L92 


-91 


-99 

r98 

^97 

r 

-96 
|-95 
7  94 
-93 


-92 


-91 


-99 


-98 


-97 


-96 


95 


-94 


-93 


-92 


■99 


-98 


-97 


-96 


-95 


•94 


-93 


-92 


li         |J- 

'2  '4 


r9i 
FULL 


-91 


t 
3 
4 


-99 


r98 


-97 


-96 


-95 


-94 


-93 


-92 


'91 


-90 


-89 


99 
r98 
=-97 

96 
|-95 
L94 

93 
f-92 
i-91 
1-90 
i-89 
1-88 
^87 

i-86 
1-85 
-84 
■83 


.1% 

.2 

.3 

.4 

.5 

,6 

.7 

.8 

.9 

■I.Oq 

< 

O 

_i 

hl.5_^ 

_i 

u. 

I—  2 


-  2.5 


—  3 


z 

UJ 

O 


UJ 
Q. 

Z 
.3.5  - 

CO 

CO 

O 


—  4 


i  I 

2  4 

EFFICIENCY    IN    PERCENT    AT    LOADS    STATED 


—  4.5 


5% 


z 
O 


TO  OBTAIN  EFFICIENCY  AT  ANY  LOAD  LAY  STRAIGHT  EDGE  AT  GIVEN  IRON  AND  COPPER  LOSS  POINTS 
AND  READ  THE  EFFICIENCY  AT  REQUIRED  LOAD  ON  THEIR  RESPECTIVE  SCALES  WHERE  STRAIGHT  EDGE 
CROSSES  THEM. 

VICE  VERSA.  TO  OBTAIN  LOSSES.  PLACE  STRAIGHT  EDGE  ACROSS  ANY  TWO  GIVEN  EFFICIENCY  POINTS 
AND  READ  PER  CENT  IRON  AND  COPPER  LOSS  ON  THEIR  RESPECTIVE  SCALES. 


GRAPHICAL  SOLUTION  FOR  LONG  LINES 


75 


CHART  X-PETER'S  REGULATION  CHART 

FOR  DETERMINING  TRANSFORMER  REGULATION 


POWER  FACTOR  OF  LOAD  IN  PER  GENT. 
00  95  90    85    80     •   70     60 


"T 


Er~i 


I— 


—  I 


2— 


»-  H 

z 

UJ 

O 

cc' 

UJ 
Q. 


3— 


UJ 

o 
z 
< 

<o 

CO 

UJ 

tr 


4— 


—  2 


z 

UJ 

O 

'en 

UJ 
0. 

—  z 


—  3 


z 

O 

\- 

< 

■_l 

D 
O 

UJ 

cr 


—4 


l-z^ 

13 

I 

J 

-J 

-J 

I 

— I 

2-"-! 

-I 

~3 

-3 

1 

-I 

-n 

-D 

3-1 
-1 

-I 
--I 

-| 
~J 
—i 

-I 
— H 

1 
"H 

-| 

"D 
_J 

J 
_j 

J 
4-1 


Z  "1 
UJ— 1 

o  n 

UJ       ' 

<  -^ 
H  -I 
CO--I 

CO  -1 
UJ       I 

tr— I 

6-:i 

--1 

H 

--J 

J 

J 

I 

n 

"1 
—I 

L 


- 


2— 


2  — 


^3- 

UJ 

O 

UJ 


z 
o 

h4- 
< 

_i 

o 

UJ 
QC     - 


I  — 


2— 


3  — 


-3— 


4— 


4— 


2  — 


3— 


5— 


5— 


6  — 


-     7— 


H     H^ 


6— 


I— 


2  — 


3  — 


4— 


7  — 


I— 


2  — 


3— 


4- 


5— 


6  — 


7  — 


t- 
f-- 

1^2 

t: 

(-- 

P" 

r-- 
r- 

F-3 
h- 

c: 
I — 
I" 
r- 
[". 

U-4 

F- 

t- 

h- 

b" 

t- 

\- 

F- 

[-- 

^:^ 

t.  UJ 
I-    O 

c 
t 

c.  z 

I-  ~ 
c- 

1—6 

f- 
r 
Tuj 

:§ 

U     UJ 

^:| 

r- 

L- 

C-7 

h- 
C. 

^ 
r- 

r 

I— 

L 

L 

L- 

_k 


REGULATION  CHART 

TO  OBTAIN  THE  REGULATION  FOR 
A  LOAD  OF  ANY  POWER  FACTOR.  LAY 
A  STRAIGHT  EDGE  ACROSS  THE 
GIVEN  RESISTANCE  AND  REACTANCE 
VALUES  AND  READ  THE  REGULATION 
AT  REQUIRED  POWER  FACTOR  ON 
THEIR  RESPECTIVE  SCALES. 

TO  OBTAIN  THE  RESISTANCE  AND 
REACTIVE  COMPONENTS  WHEN  THE 
REGULATIONS  AT  ANY  TWO  POWER 
FACTORS  ARE  GIVEN.  PLACE  THE 
STRAIGHT  EDGE  ACROSS  THE  GIVEN 
POINT  AND  READ  THE  REQUIRED 
COMPONENTS  ON  THEIR  RESPEC- 
TIVE SCALES. 

BROKEN  LINES  ARE  FOR  DETER- 
MINING THE  REACTANCE  WHEN  THE 
IMPEDANCE  DROP  AND  RESISTANCE 
DROP  ARE  KNOWN.  THIS  IS  DETER- 
MINED  WITH  A  STRAIGHT  EDGE. 


— o  f'3 


Z 
ui 
O 

■q: 

UI 
Q. 

h  z 


cr 

UJ 
Q. 


c- 
1-- 

I 

c: 
I— 

L- 

I-  — 

F- 

(-"- 

L". 

f. 

t.. 

t- 

C" 

t- 

\- 

C- 

h 

C" 

^- 

r- 
d" 
h- 
CL 

u 

L_  UJ 

r    °- 
r-  z 


z 

UJ 

o 


I— 

I- 
I — 

r 


-    UJ 


o  r- 

r 


< 

-I- 

o 

■< 


r 

L_ 

L 


P-uj  L. 
<r  L 
L.. 
L 
-4  !- 


r 

r- 

I- 

r~ 

r 

I— 

h 


I- 

r 
u- 

u 


UI 

O 

z 
< 

O 
< 

UJ 


100  95       90  85  80       70      60 

POWER  FACTOR  OF  LOAD  IN  PER  CENT. 


76 


GRAPHICAL  SOLUTION  FOR  LONG  LINES 


the    resistance   per    conductor   including    an    equivalent 
value  to  correspond  to  the  resistance  in  the  high  and 
low  tension  windings  of  two  transformers  will  be, — 
/?  +  y^j  =  105  +  6.25  +  6.2s  =  1 17.5  ohms. 

The  percent  reactance  volts  of  a  transformer  hav- 
ing 3.74  percent  regulation  at  80  percent  lagging  power- 
factor  and  1.04  percent  resistance  volts  may  be  read 
directly  from  Peter's  Regulation  Chart  (Chart  X)  by 
laying  a  straight  edge  along  the  points  corresponding  to 
1.04  percent  resistance  and  3.74  on  the  80  percent  power- 
factor  line.  The  intersection  of  the  straight  edge  with 
the  last  solid  line  at  the  right  will  give  the  percent  react- 
ance, :=  4.85  percent. 

The  percent  reactance  volts  can  also  be  read  di- 
rectly from  the  ]\Iershon  Chart.     To  do  this,   follow 

TABLE  M— APPROXIMATION   OF  RESISTANCE~AND"rE- 

ACTANCE  VOLTS  FOR  TRANSFORMERS  OF 

VARIOUS  CAPACITIES 


ohms  reactance  will  therefore  be 


2882 


:=  28.84  ohms 


Transformer 
Capacity 
in  Kv-a 

Voltage  Drop  in  Percent 

Resistance 

Reactance 

25  cycles 

60  cycles 

25  cycles 

60  cycles 

300 
500 

750 

2.  IS 

1.4 

1.2 

1-3 
1.2 
I.I 

4.0 

41 
4.2 

5-6 
6.0 
6.3 

1000 
1500 
aono 

1-7 
1-4 
1-3 

I.I 
o.g 
0.8 

6.0 
6.2 
6.4 

6.5 
7.0 
7.0 

3000 
5000 
7500 

1.2 
I.I 
1.0 

0.75 
0.6s 
0.6 

6.8 

7.2 
7.8 

7.0 
7.0 
8.0 

10  000 
15000 
25000 

I.O 

0.9S 
0.9 

0.6 

0.55 

0.5 

8.0 
8.0 
8.0 

8.0 
8.5 
9.0 

upward  the  vertical  line  in  the  Mershon  Chart  corre- 
sponding to  80  percent  power-factor  until  it  intersects 
the  first  arc.  From  this  point  of  intersection  follow  the 
horizontal  line  to  the  right  a  distance  corresponding  to 
T.04  percent  resistance  volts.  From  this  point  thus  ob- 
tained follow  the  vertical  line  until  the  arc  repre.senting 
3.74  percent  voltage  drops  is  reached.  The  length  of 
this  vertical  line  will  be  the  percentage  reactance  volts 
of  the  transformer,  in  this  case  4.8  percent.  Of  course 
the  reactance  may,  if  desired,  be  calculated  by  following 
the  general  construction  traced  out  as  above  described 
upon  the  Mershon  chart,  but  the  chart  will  give  sufiR- 
ciently  accurate  values  for  practical  purposes. 

The  volts  necessary  to  overcome  the  reactance  of 
the  windings  of  one  of  these  transfonners  is  therefore 
found  to  be  60  046  X  0.048  =  2882  volts  to  neutral.  The 


99.92 

to  neutral  for  each  transformer.  Since  the  reactance 
of  each  line  conductor  is  249  ohms,  the  reactance  per 
conductor,  including  an  equivalent  value  to  correspond 
to  the  reactance  in  the  high  and  low  tension  windings  of 
two  transformers  will  be, — 

X  +  Xt  =  24,g  +  28.84  +  28.84  =  306.68  ohms. 
The  impedance  of  one  conductor  of  the  circuit  of 
problem  X  including  the  raising  and  lowering  trans- 
formers will  be, — 

2  =  117.S  -|-  y  306.68  ohms 
and         Y  =  (assumed  to  be  the  same  as  without  the  trans- 
formers). 

With  the  assumed  values  for  the  impedance,  the 
performance  of  the  combined  circuit  may  be  calculated 
as  though  there  were  no  transformers  in  the  circuit. 

VOLTAGE  AND  CURRENT  AT  INTERMEDIATE  POINTS  ALONG 
THE  CIRCUIT 

Thus  far  we  have  considered  the  electrical  condi- 
tion at  the  two  ends  of  a  transmission  circuit  only.  Oc- 
casionally it  may  be  desired  to  determine  the  voltage  or 
the  current  at  a  point,  or  at  various  points  along  the 
circuit.  In  Fig.  21,  graphs  of  the  voltage  and  of  the 
current  are  shown  for  points  between  the  terminals  of 
a  circuit  corresponding  to  the  condition  of  zero  load, 
and  also  of  rated  load.  The  graphs  were  plotted  by 
determining  graphically  the  voltage  and  the  current  for 
points  at  50  mile  intervals  along  this  300  mile  circuit, 
as  follows: — 

To  determine  the  conditions  250  miles  from  the 
sending-end,  (50  miles  from  the  receiving-end)  the 
three  auxiliary  constants  were  obtained  from  the 
Wilkinson  charts  corresponding  to  a  circuit  50  miles 
long.  In  other  words,  it  was  assumed  that  the  circuit 
was  only  50  miles  long.  By  multiplying  these  auxiliary 
constants  by  the  known  voltage  and  current  at  the  re- 
ceiving-end of  the  circuit,  voltage  and  current  diagrams 
were  constructed  as  in  Fig.  27  and  on  these,  the  corre- 
sponding values  of  voltage  and  current  at  the  sending- 
end  of  the  50  mile  section  were  scaled  off.  This  gives 
the  conditions,  for  the  load  assumed,  at  a  point  250 
miles  from  the  sending-end.  In  a  similar  manner  the 
voltage  and  current  at  this  point,  corresponding  to  zero 
load  at  the  receiving-end,  may  be  obtained.  A  similar 
precedure  will  determine  the  electrical  conditions  for  a 
point  100  miles  from  the  receiving-end  (200  miles  from 
the  sending-end).  The  auxiliary  constants  will  this 
time  be  read  from  the  charts,  corresponding  to  a  100 
mile  circuit,  but  the  same  receiving-end  conditions  will 
be  used,  as  before.  The  electrical  condition  for  any 
intermediate  points  along  any  smooth  line,  may  thus  be 
readily  determined. 


CHAPTER  IX 
PERFORMANCE  OF  LONG  TRANSMISSION  LINES 

(RIGOROUS  CONVERGENT  SERIES  SOLUTION) 


THE  APPROXIMATE  electrical  performance  of 
overhead  circuits  having  a  length  not  ex- 
ceeding 300  miles,  may  readily  be  determined  by 
the  use  of  the  Wilkinson  Charts  for  determining  the 
values  of  the  auxiliary  constants,  supplemented  by  vec- 
tor diagrams  representing  the  current  and  voltages  of 
the  circuits.  In  important  cases,  as  a  final  check  upon 
the  values  obtained  by  the  simple  graphical  solution,  a 
mathematical  solution  yielding  rigorous  results  should 
be  made.  If  the  circuit  is  more  than  300  miles  long,  a 
mathematical  solution  yielding  rigorous  values  will 
be  required  for  determining  the  correct  values  of  at 
least  the  auxiliary  constants. 

FORMS  OF  RIGOROUS  SOLUTIONS 

The  most  direct  method  for  determining  mathe- 
matically the  exact  performance  of  circuits  of  great 
electrical  length  is  by  the  employment  of  hyperbolic 
functions,  and  the  fundamental  equations  are  usually 
expressed  in  such  terms.  Many  engineers  have  a 
general  aversion  to  the  use  of  mathematical  expressions 
employing  hyperbolic  functions.  One  reason  for  this  is 
that  the  older  engineers  attended  college  before  the 
hyperbolic  theory  as  applied  to  transmission  circuits  had 
been  developed,  and  tables  of  such  functions  were  not  at 
that  time  available. 

In  1893  Dr.  A.  E.  Kennelly  introduced  vector 
arithmetic  into  alternating-current  computation  for  the 
first  time.*  Although  real  hyperbolic  functions  had 
well  recognized  uses  in  applied  science,  it  was  in  1894** 
that  he,  for  the  first  time,  suggested  and  illustrated  the 
application  of  vector  hyperbolic  functions  to  the  de- 
terminations of  the  electrical  performance  of  transmis- 
sion circuits.  Since  ihat  time  Dr.  Kennelly  has  been  a 
most  persistent  advocate  of  the  employment  of  these 
functions  in  electrical  engineering  problems.  To  ad- 
vance their  use,  he  has  calculated  and  published  numer- 
ous tables  and  charts  of  such  functions.  Such  tables 
were,  until  recently,  incomplete  and  the  result  was  that 
it  was  necegsary,  in  using  these  tables,  to  interpolate 
values,  thus  introducing  complications  and  inaccuracies 
into  the  calculations. 

Tables  of  hyperbolic  functions  and  charts  are  now 
sufficiently  extensive  and  complete  for  accurate  work. 
The  universities  quite  generally  are  encouraging  instruc- 
tion of  students  in  the  hyperbolic  theory.     It  is  there- 


*Trans.  Am.   Inst.  Elec.  Engrs.,  Vol.   X,  page   175  "Im- 
pedance." 

~  **"Elecfrical  World",  Vol.  XXIII,  No.  i,  page  17,  January 
1894,  "The  Fall  of  Pressure  in  Long-Distance  Alternating- 
Current  Conductors." 


fore  to  be  expected  that,  in  the  future,  the  employment 
of  hyperbolic  functions  for  the  solution  of  long  trans- 
mission lines  will  come  into  general  use. 

The  fundamental  hyperbolic  equations  expressing 
the  electrical  behavior  of  transmission  circuits  may  be 
expressed  in  the  form  of  convergent  series  and,  in  such 
form  have,  in  some  cases,  certain  advantages  over  the 
hyperbolic  form.  The  convergent  series  form  of  solu- 
tion does  not  require  the  employment  of  tables  or  charts 
of  hyperbolic  functions,  whereas  hyperbolic  forms  of  so- 
lutions do  require  such  tables  or  charts.  If,  therefore, 
such  tables  or  charts  are  not  available,  hyperbolic  solu- 
tions cannot  be  employed. 

While  the  amount  of  arithmetical  work  involved  is 
considerable,  any  degree  of  accuracy  may  readily  be  ob- 
tained by  the  convergent  series  solution  by  working  out 
the  terms  for  the  auxiliary  constants  until  they  become 
too  small  to  have  any  effect  upon  the  results.  This  can 
also  be  done  with  hyperbolic  functions,  but  exact  inter- 
polation of  such  functions  from  tabular  values,  may  be 
considered  more  difficult  than  the  working  out  of  an 
extra  term  or  two  in  the  convergent  series  form  of  solu- 
tion. The  above  remarks  apply  to  cases  where  an  un- 
usual degree  of  accuracy  is  required.  Later  will  be  in- 
cluded a  tabulation  of  the  performance  of  64  different 
electrical  circuits,  as  determined  by  a  rigorous,  and  also 
by  eight  different  approximate  methods  of  calculation. 
As  the  rigorous  values  are  taken  as  100  percent  correct, 
in  determining  the  percent  error  by  the  approximate 
methods,  it  was  important  that  the  so  called  "rigorous" 
values  be  exact.  To  make  them  so,  it  was  found  con- 
venient to  employ  the  convergent  series  form  of  solu- 
tion for  these  particular  problems,  covering  circuits  up 
to  500  miles  long  and  potentials  up  to  200000  volts. 
For  the  calculation  of  the  performance  of  practical 
power  transmission  circuits,  tables  of  hyperbolic  func- 
tions are  now  sufficiently  complete  to  yield  results  well 
within  the  errors  due  to  variation  in  the  assumed  linear 
constants  of  the  circuits  from  their  actual  values. 

The  employment  of  convergent  series  requires  a 
working  knowledge  of  complex  quantities  only,  whereas 
the  employment  of  hyperbolic  functions  in  addition 
leads  into  hyperbolic  trigonometry.  As  literature  per- 
taining to  the  hyperbolic  theory  becomes  more  generally 
available,  and  as  the  younger  engineers  take  up  active 
engineering  work,  the  hyperbolic  theory  will  become 
more  generally  used. 

For  the  purpose  of  providing  a  choice  of  rigorous 
methods,  both  convergent  series  and  two  forms  of  hy- 


78 


CONVERGENT  SERIES  SOLUTION  FOR  LONG  LINES 


perbolic  solutions  are  given.  The  numerical  values 
employed  in  these  solutions  have  been  carried  to  what 
may  appear  as  an  unnecessary  degree  of  precision.  The 
reason  for  this  is  to  demonstrate  the  fact  that  all  of 
these  rigorous  solutions  yield  the  same  results.  For 
practical  problems  less  accuracy  would  be  essential, 
thus  reducing  the  amount  of  arithmetical  work. 

Before  taking  up  the  rigorous  solutions,  it  has  been 
thought  desirable  to  review  the  rules  regarding  the  use 
of  complex  quantities  and  vector  operations. 

COMPLEX    QUANTITIES 

The  calculation  of  the  auxiliary  constants  of  the 
circuit  by  convergent  series,  and  the  further  calculation 
of  the  electrical  performance  of  the  circuit,  involve  the 
use  of  complex  numbers,  that  is,  numbers  containing  / 
terms.  Thus  A  ^=  a^  -\-  ja^  is  a  complex  quantity.  To 
the  beginner,  expressions  containing  /  terms  may  seem 
difificult  to  understand.  It  cannot  be  made  too  emphatic 
that  the  rules  governing  the  use  of  such  terms  are  so 
simple  (embodying  only  the  simple  rules  of  algebra) 
that  the  beginner  will  shortly  be  surprised  with  the  ease 
at  which  complex  quantities  are  handled. 

y  Terms — In  the  complex  notation  Z  =^  X  -\-  jY, 
the  prefix  /  indicates  that  the  value  Y  is  measured  along 
the  axis  perpendicular  to  that  of  X,  or  what  is  called 
the  imaginary  axis.  There  need  be  no  significance  at- 
tached to  the  symbol  /  other  than  that  of  a  mere  dis- 
tinguishing mark,  to  designate  a  distance  above  or  be- 
low the  reference  axis  in  the  vector  diagram.  How- 
ever, great  use  is  made  of  a  further  assigned  signifi- 
cance. It  has  a  numerical  significance  in  the  form  of 
j  =  1  w  which  enables  all  formal  algebraic  opera- 
tions, multiplication,  addition,  extraction  of  roots,  etc. 
incident  to  computation  involving  complex  quantities,  to 
be  carried  out  rigorously.  This  numerical  designation 
for  y  does  not  prevent  its  use  as  a  designating  symbol 
for  the  vertical  direction  in  the  vector  diagram.* 

PLANE  VECTORS 

Alternating  voltages  and  currents  which  vary  ac- 
cording to  the  sine  or  cosine  law,  may  be  represented 
graphically  by  directed  straight  lines,  called  plane  vec- 
tors. The  length  of  the  vector  represents  the  effective 
value  of  the  alternating  quantity,  while  the  position  of 
the  vector  with  respect  to  a  selected  reference  vector, 
base  or  axis,  gives  the  phase  displacement.  The  line 
OP,  of  Fig.  29,  represents  a  plane  vector  inclined  at  an 
angle  of  33°  41'  with  the  base  OS  (the  axis  of  refer- 
ence). The  length  of  the  line  OP  is  a  measure,  to  some 
assumed  scale,  of  the  effective  value  of  the  voltage  or 
current,  while  the  angle  SOP  gives  the  phase  displace- 
ment. 

Counter-clockwise  rotation  is  considered  positive. 
Thus,  in  Fig.  29,  if  the  line  OS  represents  the  instantan- 
eous direction  of  the  current  and  the  line  OP  that  of  the 
voltage  at  the  same  instant,  the  current  is  represented 


♦For  an  extended  explanation  of  /  terms,  reference  is  made 
to  Dr.  Charles  P.  Steinmetz's  "Enginering  Mathematics",  and 
Dr.  A.  E.  Kennelly's  "Artificial  Electric  Lines." 


as  lagging  behind  the  voltage  by  the  angle  33°  41',  By 
m.eans  of  vectors  the  relative  phase  position  and  value 
of  either  currents  or  e.m.f.'s  can  be  represented  in  the 
same  manner  as  forces  in  mechanics. 

The  position  of  P,  with  respect  to  O,  is  usually  de- 
fined in  terms  of  rectangular  or  polar  co-ordinates.  In 
rectangular  co-ordinates  there  are  two  fixed  mutually 
perpendicular  axes,  ~XOX  and  -—YOY  (Fig.  31)  in 
the  plane  of  reference.  The  former,  — XOX,  is  called 
the  real  axis,  or  axis  of  real  quantities.  The  latter, 
— YOY,  is  called  the  imaginary  axis,  or  axis  of  imagin- 
ary quantities.  The  qualifying  adjective  "imaginary" 
does  not  mean  that  there  is  anything  indeterminate  or 
fictitious  about  this  axis.  The  perpendicular  projections 
of  P-i  (Fig.  31)  on  the  X  and  Y  axes  are  respectively 
the  real  component  X,  and  the  imaginary  component  Y. 

The  magnitude  and  sign  of  the  rectangular  com- 
ponents X  and  Y  completely  determine  the  position  of 
the  vector  OP.  Positive  is  indicated  to  right  and  up- 
ward, negative  to  the  left  and  downward  as  indicated  in 
Fig.  30.  Thus,  if  X  and  Y  are  both  positive,  OP  lies  in 
the  first  quadrant.  If  X  and  Y  are  both  negative,  OP 
lies  in  the  third  quadrant.  If  Z  is  —  and  Y  is  +,  OP 
lies  in  the  second  quadrant.  If  X  is  +  and  Y  is  — ,  OP 
lies  in  the  fourth  quadrant.  Any  plane  vector  may  be 
completely  specified  by  its  real  and  imaginary  compon- 
ents X  and  Y.  Thus,  beneath  Fig.  31,  is  a  table  in 
which  the  point  P  is  located  in  the  plane  by  co-ordinates 
for  all  quadrants. 

From  Fig.  30  it  is  evident  that,  mathematically,  the 
quadrature  numbers  are  just  as  real  as  the  others.  The 
quadrature  numbers  represent  the  vertical,  and  the  ordi- 
nary numbers  the  horizontal  directions. 

VECTOR  OPERATIONS 

In  general,  in  the  handling  of  complex  numbers  in- 
volving y  terms,  the  simple  rules  of  algebra  are  followed. 
In  Fig.  32  two  vector  quantities  are  shown.  Vector  A 
has  a  magnitude  of  5  units  and  is  inclined  in  the  positive 
or  leading  direction  at  an  angle  of  36°  52'  with  the 
horizontal  reference  vector,  and  vector  B  has  a  magni- 
tude of  4.47  units,  and  is  inclined  in  the  positive  or 
leading  direction  at  an  angle  of  63°  26'  with  the  refer- 
ence vector.  These  vector  quantities  are  expressed  in 
rectangular  co-ordinate  rs  A  ^=  -\-  4  -\-  jj,  B  =  -\-  £  -\- 
j/j  or  in  polar  co-ordinates  as  ^  =  5  /jd°  52',  B  =  4.47 
/63°  z6'.  The  prefix  j  simply  means  that  the  number 
following  it  is  measured  along  the  vertical  or  Y  axis. 
The  dot  under  the  vector  designation  indicates  that  A  is 
expressed  as  a  complex  number,  so  that  the  absolute 
value  of  A  would  be  V  i.4)^  -\'  (j)*  =  5  ^"d  of  B  = 
V  {sY  -\-  {4Y  ^=  4.47.  The  absolute  value  of  a  com- 
plex number  is  called  its  "size" ;  while  the  angle  is  called 
its  "slope". 

In  order  to  illustrate  the  handling  of  complex  quan- 
tities, the  various  operations  of  addition,  subtraction, 
multiplication,  division,  evolution  and  involution  of 
the  vectors  A  and  B  in  Fig.  32,  will  be  performed. 


CONVERGENT  SERIES  SOLUTION  FOR  LONG  LINES 


» 


Addition — Fig.  33  illustrates  the  addition  of  these     units  and  is  inclined  in  the  forward  direction  at  a  slope 
vectors  expressed  in  rectangular  co-ordinates.    The  re-     of  49°  24'  with  reference  to  the  initial  line,  OS. 
suiting  vector  will  have  as  its  real  component,  the  alge-  Subtraction — Fig.  34  illustrates  the  subtraction  A  — 


■^«^ 


>#^    33- »^' 


\         ';       VECTOR  OF      . 
"""■"S"  REFERENCE 


FIG.  29 


FIG. 

31 

V 

■  +  j3 

p 

-a 

+j2 

P 

-1 

/ 

+  jl 

-3 

-2      -1 

0 

+  1 

1 

+2 

+  3 

->X 

-J" 

P-3 

-J2 

P-4 

-is 

-Y 

p-1    5    x-*3,    Y-+2  .•.P-I-+3+J2 

P-2   :   X--3.    Y-*2  ,".P-2--3*j2 

P-3    }    X--3.    Y--2  /.  P-3--3-J2 

p_4    ;    X-+3     Y--2  .•.P-4-+3-J2 


ADDITION    A+B 


+ 

^i3 

FIG.  30 

+  j2 

\ 

-3          -2           H 

0 

♦1 

•?     *?  ! 

-jl 

-j2 

-J3 

MULTIPLICATION 
AXB 


-J 


FIG.  32 

VECTORS  A  &,   B 


VECTOR  (A) -(a| +182)= +4+]  3-  S /if  62' 
VECTOR  (B)  =(b|  +  j  b2)"*-2  +j4=  AA7  /S3i2Sl 


SUBTRACTION    A-B 


FIG.  34 


(A)=  +4*j3 
-(B)-  -  2  -  j  4  -CHANGE  SIGNS  AND  ADO. 
(C)'<-2-il 


<C).lf(2)2+(l)2  -  2 J4 /-?y?4' 


DIVISION 


B 


(A)=  +  4  +  j3 
+  (B)-+2  +  i4 


((J)=«.e  +  j7 

(C)=Y(6)*+m^  -  9.22  /4y24' 


FIG.  35       "N, 

O-VL-tlS 

VECTOR  (C)  =  I .  I  2  -^^^f  -  26*  34- 

A 

LENGTH   VECTOR  (C)—g-=  I  12 
ANGLE    VECTOR  (C)  -      " 

36*  52 -63'  26'  "  "  26°  34' 


{l^)-->t*j3 
(g|=»2<-)4 

-12-t-ie 

(C)=-4+i22 


FIG.  36 


<C) 


.YW 


4.(22)'  -  22.36 
OR 


/'go  "If 


LENGTH  (0=  &  X  4  47  -  22.36 
ANGLE     (C)«=»3*26'*38'52'-I00-|8 


FIGS.    29   TO   36 — EXAMPLES    OF   VECTOR    SOLUTIONS 


braic  sum  of  the  reals,  and  as  its  imaginary  component,  B.     This  is  simply  addition  after  the  signs  of  both  of 

the  algebraic  siun  of  the  imaginaries.    Thus :  the  components  of  the  vector  to  be  subtracted  have  been 

A  =  +  4  +  J3  reversed.     Thus,— 
+  B  =  +  2  +  14 


A+B  =  C=  +  6  +  }7 


C  =  V  (6)'  +  (7)'  =  9.22  absolute. 
The  resulting  vector  has,  therefore,  a  size  of  9.22 


A  =  +  4  +  J3 
—  B  =  —  2  —  /4 
A  -  B  ='C  =  +  2  —  ii 


C  —  y  (2)'  +  (i)'  =  2.24  absolute. 


CONVERGENT  SERIES  SOLUTION  FOR   LONG  LINES 


The  resulting  vector  C  has  therefore  a  size  of  2.24 
units  and  a  slope  of  —  26°  34'.     In  polar  co-ordinates, 

C  =  2.24\26°   j/. 

Division — To  divide  one  plane  vector  by  another, 
divide  their  sizes  and  subtract  their  slopes.  Fig.  35. 
Thus, — 


Absolute  value  of  C  : 


1. 12 


4-47 
Angle  of  inclination  of   C  =  36°   52'  —  63°  2&  = 

—  26°  34'  in  the  negative  direction.  In  polar  co-ordi- 
nates C  =^  i.i2\26°  S4'- 

Multiplication — Fig.   36  illustrates  the  multiplica- 
tion of  the  vectors  A  and  B.     Here  the  rules  of  algebra 

also  apply,  except  that  when  two  /  terms  are  multiplied 
signs  are  assigned  opposite  to  those  which  would  be  used 
in  the  ordinary  solution  of  an  algebraic  problem.  This 
is  for  the  reason  that, — 

hence,  p  ■=  —  / 

Hence  where  p  occurs  it  is  replaced  by  its  value 

— I  and  therefore, — 

-  ;■  X  ;■  =  -f  7 
;■•  =  —  ;■ 
)*  =  +  I 
i'  =  -\-  j,  etc. 
Thus,  to  get  the  product  of  A  and  B : — 

^  =  +  4  -1-  ;■  3 
B  =  +  2  -F  y  4 

+  8  -f  y  6 
—  12  -I-  y  16 

^X5  =  C  =  —  4-fy22  =  22.35  absolute 
The  resulting  vector  C  has  therefore  a  size  of  22.35 
units  and  is  inclined  in  the  positive  direction  at  an  angle 
of  100°  18'  to  the  vector  of  reference.     The  polar  ex- 
pression is  C  =  22.35  \  100°  18' 

The  magnitude  and  position  of  the  product  may  be 
also  determined  by  multiplying  the  sizes  of  the  vectors 
and  adding  their  slopes.     Thus : — 

Size  of  C  =  5  X  447  =  22.35  (as  above) 
Slope  of  C  =  63°  26'  -1-  36°  52'  =  100°  18'. 

Involution — Involution  is  multiple  multiplication. 
To  obtain  the  power  of  a  plane  vector,  find  the  power  of 
the  polar  value  and  multiply  the  angle  by  the  power  to 
which  the  vector  is  to  be  raised.  Thus, — vector  A  =^ 
5  /36°  5^:  and  (5  /jd"  s^y  =  f  /yf  44'  =  25 
/73°  44'- 

Evolution — To  find  the  root  of  a  polar  plane  vector, 
find  the  root  of  the  polar  value  and  then  divide  the 
slope  by  the  root  desired.  Thus  vector  A  =  5  /j6°  52'; 
and  V  5  /j(5o  ^2'  =  2.236  /j8°  2&. 

SOLUTION    BY   CONVERGENT   SERIES 

The  hyperbolic  formula  for  determining  the  operat- 
ing characteristics  of  a  transmission  circuit  in  which 
exact  account  is  taken  of  all  the  electric  properties  of 
the  circuit  is  frequently  expressed  in  the  following 
form, — 


E^=E,  cosh  VZY  +  h  yj^  sink  ^ ZV (5/) 

U  =  h  cosh   VZY  +  Ei      L    ^■"'''  ^/^■ (52) 

Since  yZY  is  complex,  the  hyperbolic  functions  of 
complex  quantities  are  required  in  solving  these  equa- 
tions. 

In  above  formula,  expressed  in  hyperbolic  language, 
the  three  auxiliary  constants  A,  B  and  C  which  take  into 
account  the  "distributed"  nature  of  the  circuit  are  repre- 
sented by  the  quantities — 

A  =  Cosh  yZK (53) 

B  =   yj~    sinh  yjlY (54) 

C  =  — (=r      sink  \ZY. (55) 

Vf.    . 

Equations  (51)  and  (52)  above  may  therefore  be 
expressed  in  terms  of  the  auxiliary  constants.  A,  B  and 
C,  as  follows: — 

E.  =  Er  A  +  L  B (5(5) 

I,.=  h  A  +  Er  C  (57) 

or    Er  =  E.  A  —  I.  B (58) 

I,  ^  I.  A  —  E.C (5P) 

These  three  auxiliary  constants  may  be  calculated 
by  convergent  series  as  follows : — 

r  YZ       Y'^Z^      Y^Z^      Y'Z*  1 

[YZ       V^  Z^       Y^  Z^        Y*  Z*  1 

^  T  YZ       Y^Z^       Y^Z»        Y*Z*  1 

C=  y  [/  +^-  +-j^  +^-j^  +  J^J^+  etc.\  ...  (6,) 

The  above  series  are  simply  expressions  for  the 
auxiliary  constants  as  previously  stated.  These  con- 
stants are  functions  of  the  physical  properties  of  the 
circuit  and  of  the  frequency  only,  and  not  of  the  volt- 
age or  the  current.  After  the  values  for  the  auxiliary 
constants  have  been  calculated  for  a  given  circuit  and 
frequency  their  numerical  values  may  be  applied  di- 
rectly to  any  numerical  values  of  E  and  /  for  which  a 
solution  is  desired.  From  this  point  on,  the  perform- 
ance of  the  circuit  may  be  determined  either  by  the 
graphical  method  previously  described  or  by  mathe- 
matical calculation. 

Any  degree  of  accuracy  may  be  obtained  by  the  use 
of  convergent  series  for  determining  the  auxiliary  con- 
stants, by  simply  using  a  sufficient  number  of  terms  in 
the  series.  The  rapidity  of  convergence  of  these  series 
is  dependent  upon  the  value  of  the  argument  ZY  and 
thus  upon  the  square  of  the  length  of  the  circuit  and 
frequency,  and  also,  to  a  lesser  extent  upon  the  product 
of  total  circuit  conductance  and  total  circuit  resistance. 


CON  V URGENT  SERIES  SOLUTION   FOR   LONG   LINES 


8i 


As  far  as  calculations  based  upon  the  more  or  !es« 
uncertain  values  of  the  fundamental  constants  of  the 
circuit  are  concerned,  the  use  of  three  terms  in  the 
series  expression  yields  results  in  a  300  mile  circuit 
which  are  sufficiently  close  to  the  exact  values  as  given 
by  the  use  of  hyperbolic  functions  (infinite  number  of 
terms).  In  the  case  of  shorter  circuits  two  terms  will 
give  a  high  degree  of  accuracy.  The  number  of  terms 
necessary  will  be  determined  while  doing  the  work,  for 
it  is  usual  to  figure  out  the  terras  of  the  series  until  they 

become  too  small  to  be  considered  when  added  to  — - 

3 

V  z 
or-^. 

In  Table  N  are  given  values  for  the  auxiliary  con- 
stants (expressed  in  rectangular  co-ordinates)  illustrat- 
ing the  convergence  of  the  series  for  a  300  mile,  60 
cycle  circuit  (Problem  X),  the  complete  calculation  of 
which  will  follow. 

Table  A''  shows  that  even  for  a  60  cycle,  300  mile 
circuit,  three  terms  give  sufficiently  accurate  results 
for  determining  constant  A,  whereas  two  terms  are 
sufficient  for  determining  constants  B  and  C.  This  is 
on  account  of  the  slower  convergence  of  the  hyperbolic 
cosine  series. 

TABLE  N— CONVERGENT  SERIES   TERMS  FOR  PROBLEM  X. 


No.  of 
Terms 

Constant  A 

Constant  B 

Constant  C 

I 
2 

3 

4 

Infinite 

i.oooooo  +  j  0.000000 
+  0.805407  -I-  j  0.082057 
-I-  0.810596  +  j  0.07673s 
+  0.810558  +  j  0.076832 
+  0.810558 +  j  0.076831 

105  +  j  249 
+  91.3788 +  j  235.7211 
+  91.7527  +  j  235.8678 
+  91.7486  +  j  235.8680 
+  91.7486 +  j  235.8680 

0             +  j  0.001563 

—  0.000043  +  j  0.001462 

—  0.000041  +  j  0.001463 

—  0.000041  +  j  0.001463 

—  0.000041  +  j  0.001463 

CALCULATION   FOR  THE  AUXILIARY   CONSTANTS  BY 
CONVERGENT  SERIES 

The  form  of  solution  and  procedure  indicated  in 
Chart  XI  for  the  calculation  of  the  auxiliary  constants 
by  convergent  series  is  suggested  as  being  complete  and 
easy  to  follow. 

First  the  physical  characteristics  of  the  circuit  and 
the  frequency  are  stated.  These  are  the  only  features 
having  any  bearing  upon  the  value  of  the  auxiliary  con- 
stants for  a  given  circuit.  The  voltage  and  current  to 
be  transmitted  do  not  affect  the^e  constants.  The  re- 
sistance, reactance,  conductance,  and  susceptance  to 
neutral  per  mile  are  ascertained  from  the  tables  for  one 
conductor  of  the  circuit.  These  values  are  then  multi- 
plied by  the  length  of  the  circuit  in  miles  and  set  down 
as  total  per  conductor. 

The  values  of  Y  and  Z  must  now  be  set  down  for 
the  problem  in  the  form  of  complex  quantities.  Thus 
Z  =  7?  -f  ;X  =  705  -f  ;>./p  and  F  =  G  -f  /B  =  0  + 
/o.oo/5<5?  since  zero  leakage  conductance  has  been  as- 
sumed for  this  case.  Conductance  G  represents  the 
true  power  loss  in  the  form  of  leakage  over  insulators 
and  of  corona  loss  through  the  air  between  conductors. 
Corona  loss  corresponding  to  the  assumed  atmospheric 
conditions  may  be  estimated  by  applying  Peek's 
formula     (See    Chapter    IV    on    Corona).     Insulator 


leakage  may  be  approximated  from  the  most  suitable 
test  data  available.  It  is  general  practice  in  the  solu- 
tion of  all  but  the  very  longest  high-voltage  circuits  to 
ignore  the  effect  of  the  losses  due  10  leakage  and  corona 
effect.  These  losses  will  be  ignored  in  this  case,  so  that 
C  becomes  zero.  After  Z  and  Y  have  been  written 
down  in  the  form  of  complex  quantities  the  product  YZ 
should  be  found  as  previously  described  for  the  multi- 
plication of  complex  quantities.  The  second,  third  and 
fourth  power  of  YZ  may  then  te  found,  if  desired. 
Chart  XI  shows  the  fourth  power,  bur  on  ali  bur  the 
longest  circuits  a  total  of  four  terms  will  be  sufficient, 
and  for  most  problems  three  terms  will  give  sufficient 
accuracy.  The  range  of  ciccuracy  has  been  previously 
indicated  for  a  300  mile  circuit  on  the  basis  of  any 
number  of  terms  being  used  up  to  and  including  in- 
finity. The  values  in  Chart  XI  are  carried  out  to  six 
decimal  places  whereas  four  places  will  usually  give 
sufficient  accuracy  ^"r  calcf'-ifinfr  th*!  "tlue*  <^i  th<»  i-""- 
stants  A  and  B.  The  smallness  of  the  value  of  con- 
stant C  may  make  six  places  desirable  when  calculating 
its  value. 

After  the  values  of  YZ,  Y^  Z^,  1"  Z*  etc.,  have 
been  calculated  they  are  divided  by  2,  24,  720  etc.,  re- 
spectively, set  down  and  added  to  i.  This 
gives  the  value  of  the  auxiliary  constant 
A,  as  -f-  0.810558  +  ;  0.076831  which  is 
also  referred  to  as  a^  -j-  /oj.  The  abso- 
lute value  of  the  constant  A  =  0.8142  is 
simply  the  square  root  of  the  sum  of  the 
square  of  a^  and  o^.  The  polar  value  of 
A  is  thus  0.8142  /5°  24'  53". 
The  solution  for  the  constant  B  is  of  the  same 
general  form  as  the  solution  for  the  constant  A,  except 
that  the  values  of  YZ,  Y^*Z^,  and  7'  Z'  etc.,  are  divided 
by  6,  120  and  5040  respectively.  After  these  results 
are  added  to  /  they  are  multiplied  by  Z,  the  product  be- 
ing the  value  of  the  auxiliary  constant  B  ox  h^  -\-  ;'&,. 
The  absolute  value  of  B  is  obtained  in  the  same  manner 
as  the  absolute  value  of  A. 

The  solution  for  C  is  the  same  as  for  B  except  that 
in  place  of  the  constant  B  series  being  multiplied  by  Z 
it  is  multiplied  by  Y  and  the  values  of  C  or  c,  +  /c, 
obtained. 

AUXILIARY   CONSTANTS   OF   VARIOUS    CIRCUITS 

In  Chart  XII  are  tabulated  exact  values  for  the 
auxiliary  constants  for  the  64  problems  to  which  fre- 
quent reference  will  be  made.  These  auxiliary  con- 
stants have  been  calculated  by  convergent  series,  the  re- 
sults having  been  checked  through  the  medium  of  three 
separate  calculations  made  at  different  times.  They 
are  therefore  believed  exact  to  at  least  five  significant 
digits.  The  results  have  been  expressed  in  both  rec- 
tangular and  polar  co-ordinates. 

CALCULATIONS   OF    PERFORMANCE 

In  Chart  XIII  is  given  the  complete  calculation  of 
the  electrical  performance  for  problem  X,  starting  with 


82 


CONVERGENT  SERIES  SOLUTION  FOR  LONG  LINES 


CHART  XI— EXAMPLE  ILLUSTRATING  RIGOROUS  SOLUTION   FOR  THE  AUXILIARY  CONSTANTS 

BY  CONVERGENT  SERIES  FOR  PROBLEM  X. 

PHYSICAL  CHARACTERISTICS 


OF  CIRCUIT  -FREQUENCY 

LENGTH^  300  MILES.  CYCLES,  60. 

CONDUCTORS  — <t  000  STRANDED  COPPER. 
SPACING  OF  CONDUCTORS  10  X  10  X  20  FEET. 
EQUIVALENT  DELTA  SPACING  =  l2.e  FEET. 

LINEAR    LINE    CONSTANTS 

FROM  TABLES  -  PER  MILE 

TABLE    NO.  2,  r  =  .350  OHM  AT  25°  C, 
TABLE    NO.  5,X  =  .830  OHM  (BY  INTERPOLATION). 
TABLE  NO.  10  b=  5.21  i(  10"^  MHO  (BY  INTERPOLATION). 
g=(IN  THIS  CASE  TAKEN  AS  ZERO). 

TOTAL  PER  CONDUCTOR 
R=t1  =  .350  X  300  =  105  OHMS  TOTAL  RESISTANCE. 
X=*l  =  '830  X  300  =  249  OHMS  TOTAL  REACTANCE. 
B=bl  =  5.2l  X  300  X  10"^  =,001563  MHO  TOTAL  SUSCEPTANCE. 
Q=gl=     0  X  300  =  0  MHO  TOTAL  CONDUCTANCE. 


MULTIPLICATION    OF    YZ 


v 

z 

= 

0         +, 
105      t- J 

,001563 
249 

- 

0 

.  389  1  87  +  J 

0 

. 1 64 II  5 

YZ 
YZ 

- 

.389  187  + J 
.  389  187  + J 

.  1 64 1 1 5 
.164115 

+ 

.  151466  - 
.  026934  - 

.063871 
.063871 

y2z2 
YZ 

+ 

.   124532  - 
.  389187  + J 

. 127742 
,  1641 15 

+ 

.  048466  +  J 
.  020964  +j 

.  020437 
. 0497 1 5 

y3z3 
YZ 

_ 

.   027502 +  j 
.  389187  + j 

.070152 
.  1641 15 

+ 

,  0  1  0703  -  j 
.011513-] 

.  0045 1 3 
.  027302 

v*z*  = 


,  0008  I  0  ■ 


.031815 


NOTE 


THE    AUXILIARY  CONSTANTS   OF 
THE  CIRCUIT    (A)  (B)*,  (C)  MAY 
BE  OBTAINED  GRAPHICALLY 
FROM    THE    WILKINSON    CHARTS 


(A)= 

SOLUTION    FOR  (A) 

- 1^     2.       24     ''   720    '  40.320tJ 

1  .000  000 

"Y    =    -    .194593  + j  .082057 

2   2 
^^-|-    =    +   .005189- j  .005322 

y3,3 

■J—-    =    -    .000038 +j  .000097 

^i^^    =    -    .  000000 -j.  000001 

(A)=   +    .810558  + j   .076831 

(ai+jaj) 

=       0.8142 /5' 24' 53" 

(B) 

SOLUTION    FOR(B) 

-Zr  +  ^+v2z2^Y3z3       Y^Z^I 

L.         6   '    120    '  5.040 '^362,880* 

1  .  000  000 

■^    =   -    .  064863 +j  .027352 

^^    =   +   .001038- j  .001064 

y3,3 

^qIq    =    -    . 000005  +j  .000014 

.^l.n   -    -    .000000      J.OOPOOO 

(B)=  Z(+    .93617     +j. 026302) 

Z  =               1  05  +  j  249 

+      98.2978 +j  233.1063 

6.5492  +j       2.7617 

(B)=    +      91.7486 +j  235.8680 

(b|  +  j  b2) 

=  253.083  /68°  44'  4  1    OHMS 

(C) 

SOLUTION    FOR  (C) 

.,r    YZ^y2z2,y3z3,      Y-^Z^    1 
6  ^    120       6  040     362,880^ 

(C)='>'(+      93617     +j     026302) 

Y  =            0         +j     001563 

(C)=    -     000041     +j     001463 

'^'=                     (Ci+jCj) 

=.001464    \  91°  36   18"  MHO 

the  values  for  the  auxiliary  constants  and  the  receiving 
end  load  conditions  known.  The  calculations  are  car- 
ried out  by  the  employment  of  complex  numbers,  the 
complete  performance  being  calculated  for  both  load 
and  zero  load  conditions.  In  order  to  give  a  more  clear 
understanding  of  these  mathematical  operations  the 
reader  is  referred  to  the  vector  diagrams  of  Fig.  37. 

In  Chart  XIII  are  given  the  formulas  for  determin- 
ing the  E,  and  /,  values  under  load  conditions.  On 
^'&-  37  these  two  same  formulas  are  given,  but  in  the 
form  of  vector  diagrams,  upon  which  vectors  the 
numerical  values  corresponding  to  problem  X  are  stated. 
With  the  numerical  values  of  the  vectors  and  angles 
stated,  it  should  be  a  comparatively  simple  manner  to 


follow  graphically  (Fig.  37)  the  mathematical  calcula- 
tions shown  in  Chart  XIII. 

The  formulas  for  £s  and  7s  which  are  stated  in 
Chart  XIII  and  in  Fig.  37  contain  a  complex  number 
(Cos  ^r  ±  /  sin  ^r)  not  previously  stated  in  connection 
with  the  fundamental  hyperbolic  formulas  for  long  cir- 
cuits. The  formulas  previously  given  were  based  upon 
unity  power-factor.  The  introduction  of  this  new  com- 
plex number  is  made  necessary  in  order  that  the  effect 
of  the  power-factor  of  the  load  current  may  be  included 
m  the  calculations.  The  function  of  this  new  complex 
nuniii)er  is  to  rotate  the  current  vector  through  an  angle 
corresponding  to  the  power-factor  of  the  load  current. 
It  will  be  referred  to  as  the  rotating  triangle.     If  the 


CONVERGENT  SERIES  SOLUTION  FOR  LONG  LINES 
CHART  XII— AUXILIARY  CONSTANTS  OF  VARIOUS  CIRCUITS 


H 


6 

z 

Z 

UJ 

i' 

n 

Z  3 

UJO 
-■  o: 

O 

CONDUCTORS 

< 

-J 

? 

z 

i 

LINEAR  CONSTANTS 

TOTAL  PER 
CONDUCTOR  * 

AUXILIARY   CONSTANTS   OF  CIRCUIT 

THESE  AUXILIARY  CONSTANTS  TAKE  INTO  ACCOUNT  THE  EFFECT  OF  DISTRIBUTED 
CAPACITANCE  .  THEY  HAVE  BEEN  CALCULATED  RIGOROUSLY  BY  CONVERGENT  SERIES 

rl 

FROM 
TABLE 
NO   2 

xl 

FROM 
TABLE 
NOS 
4&5 

bl 

rROM 

TABLE 
NOS 
9  A  10 

%\ 

CONSTANT  (A) 

CONSTANT  (B) 

CONSTANT  (C) 

a,                  aj 

b|                 b2 

C|                         Cj 

25CYCLES                                                          1 

/ 

2 

3 

4 

20 

M 

It 

M 
H 

3 

J.J4 

5.3« 
H 

O 
0 

,'?<?'?»4  7+  J.ooo  /5s 
=  .?'7'7847   /O»o'32* 

5.^394     +  i  5.3600 
=     7.7  on     /44'3'X7' 

°          +J.000057 
=  .Oooo.?7/yo''0'o" 

3 

3.^4 

5.36 

3  7.2 
ft 

o 
o 

.9?984  7+  j.ooo  /5S 

5.5  394     +  )  5.360O- 

=        7.70R(     /44*3'27- 

0         +3,000057 
=  .oon05-7/90»0'0» 

6 
t 

7 
8 

30 

ft 

OOOO  COf>PEIZ 

• 

4 

8.3/ 

8.5 

a'." 

0 

o 

.999666  +  J.0OO336 
=   .9996  56  /o'  no" 

8. 3092   +  J  8.4999 
=      //.886      /4S'3'>'lie 

0         +  j,oooo  91 
-.OOOOBI  /^o'o'cr 

4 

8.3  < 

8.-5 

8/.0 

o 
o 

.9996S6  +  J.00O336 
=    .999  6  56  /O'  I'  10' 

S.3083    +  J   8.4999 
^        11.986    /4S'39'/^ 

0            4-  jtfOOOO  91 

zz.ooooei  /9o*o'a' 

y 

'0 

II 

IZ 

SO 

» 

OOOO  COPl='£f^ 
H 
* 

» 

4 

♦ 

'3.85 

/4.I 

/35 

o 

0 

.999049  -k-  j. 000935 
=   .<»<»<?  04  if /o°  3' /i" 

/3.84(     +  ]  /4.0  996 
=      I9.7S7    /4S'3I'44- 

0       +  ),ooo/  35 
=  .ooo/.T^V5'o»<3'o- 

6 

'3.85 

15.1 

'25 

o 
o 

,999056,  +•  j.000866 
=  .9  99  0  56/0*2' 58* 

13.941  3+   J  15.0991 
=    2O.4  83  3/47''29'a0" 

0          +   i.oOO'25 

=  .000/J.T  /9(fio'<r 

13 
/-) 
IS 

Ik 

100 

OOOO   f.QRRER 

27.7 

3  2.2 

233 

o 
o 

.996248*  J.003224 
=  .<?'?<^P5^  /O'  n'7' 

27.6307  +  J  32./  894 
=   42.4218    /4<I'3I'29" 

0          +j,000233 
=  .0005  ,,/90»0'0- 

II 

2  7.7 

3  3.2 

226 

o 
o 

.996249+  J.003IZ6 
=  .796554 /o''/0'47' 

27.6308+  j33.'874 
=  43./84  '    /5o<>/3'/r 

0         4.  i.o  00 a  2  6 

=  .OOO226/?<'"'0'0- 

n 

/3 

n 
zo 

200 

* 

ft 

ft 

II 

3<7.2 

64.8 

4  64 

o 

.98499'+  i. 009049 
-  .9ff5033  /O'ai'SS" 

38.808  +  J64.594 
-        7^.3  56/5^9«0'/0' 

—  .00000/+ 3,0  00462 
=  .oon462/f0"7'27-' 

n 

3  9.2 

6?.2 

434 

o 
o 

.98-5009+  jiooB464 
=  .9ff3^05-o/0<'29'J'' 

39.  8084+  j  68.965- 
=        7^/34    /60''37'vr8' 

— .00000 /+J.0  00432 

=  .00  04  3  2   /9C'»7'j-4!" 

Zl 

aa 

14 

3oo 

• 
* 

63^M  PLUM, 

II 

44.' 

* 

9/.2 

74  7 

o 

o 

,966085  +  J, 0/6285 
=    .96AZ3  3/0'S7-/~ 

43./033   +   j  90.409 
-     /00./S7  /64'30-36' 

—.000004  +  j.ooo 73 9 

=  .000  7.^9  /90»/7'/0* 

Zl 

44,; 

loi 

672 

o 
o 

.96,6319  +  J.O'465o 
=  .96  6  330  /o'S3-6" 

43.1070   +   j  100.077 
=     I09.966  /«»4/'48' 

—.000003+  J.0OO664 
r.000  6A4  /9cfi/S'Zr 

is 

it 
^8 

4O0 
n 

* 

* 

6  36m  flLUM, 

n 

.5  8.8 

1  30 

92  9 

o 
o 

.940/61  +  j. 036738 
=  .94054;    /'<'37'4J- 

J6.4555  +  J  127.927 
-    /3<j.5?^    /66'il'/6- 

-.ooobos  +  i. 000909 
-.onn9a4/9o'3&l'*' 

21 

3  8.8 

/34 

896 

o 
o 

.94045Z+  J.035BI9 
=  .940  80/  //'3t'S0' 

J6.4664+  ]/3;.842 
=    /43.42,r/<6''-»8'54- 

—.0  00008  +  j.ooo  878 
=  .000  878  /90»3l'/r' 

30 
3/ 
32 

5oo 

6  3  i>f^    OLUM, 
* 

n 

73.^ 

/63 

/'60 

o 
o 

.906642  +  j.o4'299 

=  .90  7.r»T    /^'S^'/.*- 

68.928     +  3  '58.928 
=     /7.-?.J>^     /«<»33''3* 

-.oooo'6  +j.oo'/24 

Zl 

-ri.S 

Its 

* 

1  IZO 
It 

o 

0 

.907/09+  3.03988O 
=  .9079S.^/.J°3''2- 

68.9507+   j   /63.76 
=     I77.684  /67'/o'0' 

-.oooo'S  +  j.oo  1  09S 
=  .00/08.7    /9o»t  -'3  3' 

60        CYCLES 

33 
31 
35 
36 

20 

• 

OOOO  COPPC 

3 

5.54 

/2.88 

137 

o 
o 

.999/19+  j, 000379 
=  .9991  IV    /O'  I'  IE" 

5.536  75  +  i  1-2.9769 
=   l^.ni/.7    /6i»4ro' 

0            +   1.000  '37 

=  .(100137/^°"°'°' 

3 

.5.54 

'2.88 

137 

o 

o 

,999//8+  J.OOO379 
=  .9991 19  /O'l'  IS" 

5.5-36  75-  +  J  11.97^9 
=  I4.0I67    /«6»44'0' 

0         +  ..000  '  37 

=  .000/17/^'''0'0' 

37 
38 
39 

30 

OOOO  CQPPCR 

4 

ft 

8.3' 

2  0.-J 

195 

o 
o 

•  99901  1  +  j,ooo8' 

-.99  9011     /o*a'4  7" 

8.299  +  120. 3887 
=    2Z.OI4     /6TSI'i- 

0            +  j.ooo  I9S 
=  .nr,n,9</90*0-i>' 

■4 

8.3' 

20.4 

IfS 

o 
o 

.9990I 1  +  j.ooo  8  1 
=  .9980  /'    /0»a'47' 

8.299    +   ].?0.3  887 
=   ZZ.OI4      /67»5/'6- 

0           +  i,OOOl9S 
=  .ooni9i/9o*0-o- 

4/ 

^3 

•50 

m 

0t 

OOOO  COPPCR 

ft 

» 

* 

/3.85 

1 

34-0 

324 

o 
0 

.994416+  j. 002539 
=  .^<7^4  9R   /0'7'33' 

1  3.79fZ    +  j  33.9479 
=     36.64J^    /«7*52'4r 

0         *j,poq323 

=  .0  00.-»3^/90«0'0' 

6 

/3.8i 

* 

36.4 

30' 

o 

0 

.994536+  j,00208; 
=  .<?<?4.r2a/o«7'/4- 

/3.7994+  j  36.3432 
=    3».»74./«^/»  30- 

0         +  i,ooo3oo 

=  .00/,^00/90<>0'0" 

4S 

47 
4S 

/oo 

m 
H 

OOOO  COPPfR 

^7,7 

77.4 

If 

562 

o 
o 

.97832+  j,007729 
=  .9783.C      /0'Z7-/a" 

27.2996   +  i  76.9/16 
=    1^1.6,119  /70»2r36' 

-k0  0ooo/+  j, 00055  8 

=  ,000,V,T*/?0»6'"- 

// 

2  7.7 
'f 

7'7.7 

^4  2 

o 
o 

.97847+  J.007452 
=  .978498   /0«a6-/4- 

2  7.3  02    +  j  7  9.196,3 
=            83.77/70»58'J<J* 

-VO0OO0/+  j.ooo  53  8 
=  .ooo.C/(8/90»<-2r 

4'? 
J-/ 

20Q 

» 
m 
m 

3O0  M  COPPti 

* 

// 

3  9.2 

/56 

1  1  Ih 

o 
o 

.9/4/28  +  j.o2;243 
=  .9/437,S-//''/9'J'- 

36.954/+  ilSI.79/ 
=    1.16.13^  /76'/1-Z' 

-.000008*  )  .00/084 
=  .00/0  84/90''2i'.?r 

/7 

3,.2 

/6i!> 

IO-44 

o 

o 

.9 I4SZ4+  j. 0/9876 
=  .9/474o  //•/4'40' 

3^.964;+  j  H,I.S07 
=     /6S.t9    /7T6-3I' 

-»00  000 7+  J  ,00  'O  (4 
=  .00/0/4    /9ti'2S4r 

-J3 

SS 
Si 

300 

» 
• 

6  3^M  f^LUM. 

9 

• 

220 

It 

1794 

o 
o 

.8088/6+  3.03700b 
=  .909663    /Z'iVo' 

38.46,55  +  3  206.359 

-   Zt)9.9/3  /79'Z6-Zg 

-.000023  +  ].0OI  67  9 
=  .00/67ff/90»47'*' 

2f 

• 

44.1 

24  3 

Ikl4 

o 
o 

.8IO027+  i,0333O7 
=  .S/070/    /Z'ZI'M' 

38,5002+  J227.9/8 
=  Z3I.I47  /8o«J4'4J- 

-.0000/8  +  i.OOl  SIO 
=  .ao/r/o  /9o'4'*- 

S7 
60 

AOO 
H 

H 

n 

6  3&M  f1LUM» 

n 

• 

58.8 

3'4 

22IZ 

ff 

o 

0 

,67/70/   +  j  .057759 
=  .674  1  7»    /4»54'J4" 

45.8726+  J  280.04 
=      2  8  3.77/  80« 4;' JO' 

-.000044+  j.oo/958 
=  .oo/9.r9  /9/«/»'0' 

2J 

•5-8.8 

322 

2/52 

1* 

o 
o 

,672455+  i  ,056208 
=  .i:74»on  /4•4«'3r' 

45-,90/3+  J287./94 

-.000042*  j  .00/9/2 
=  .oo/«A-»  /9/*/S-Z/' 

(.1 

63 

64 

^00 

m 
0 

636M  RLUM, 

m 
/' 

/  7 

73.5 

3  90 

2785 

o 
o 

.S02772+  J  .094-790 

=  ..C/l<?ff7/    /9'34';?0- 

4«.96'4  +  J  325,247 
=    32».<?l2/8/°2*'2/" 

-.OOOO8S  +  J  ,003307 
=  .oo3:^o<»  f?.?*«'3a' 

21 

7  3.  J 

402 

2690 

o 
o 

..5  04  852+  j  ,0  8'  9  69 

=  .,-r//4  6  3  /9'  I3-'Z- 

49.06/  +   j  335.4/4 
=    3  3».9»    /8''40W3- 

-.000079*  J  .002a  3o 

=  .002  3T2/92'''4.J-- 

*rl  is  the  resistance  in  ohms  at  25°  C  (77°  F),  xl  the  reacfcince  in  ohms,  bl  the  susceptance  in  micromhos  to  neutral  (multi- 
ply by  10-'  to  convert  to  mhos).  The  x  and  b  values  for  the  636 ocm  circ.  mil  aluminum  cable  were  taken  as  those  of  700000 
circ.  mil  copper  on  the  assumption  that  these  two  conductors  would  have  approximately  the  same  diameter,  gl,  the  loss  restilt- 
ing  from  leakage  over  insulators  and  from  corona  has,  for  simplicity,  been  assumed  as  zero. 


84 


CONVERGENT  SERIES  SOLUTION  FOR   LONG   LINES 


CHART  XIII— RIGOROUS  CALBULATION  OF   PERFORMANCE  WHEN  RECEIVING  END  CONDITIONS 

ARE    FIXED 


Er=I04  000   VOLTS   3  PHASE. 


KV-Ar^I  8  000.  .KWr^  I  6  200. 

PER  PHASE  TO  NEUTRAL 

KV-ApM=!^  =  6  000.     KWrn=!^  =  6  400.     Erm=^^^=60  046. 

AUXILIARY  CONSTANTS   OF  CIRCUIT 

(  A)  =- -810558 +  J.07683I  (  B)= -91.7486  *  j  235.868 

(ai->ja2)  =  (b|  +  j  b2) 

=      .8142  /s^  24'  63'  =  253.083    /68'44'  41'  OHMS 


PFr  =  90.00%  LAGGING  . 


I„  = 


6  OOP  X  I  OOP 


=  99.92  AMPERES. 


SOLUTION    FOR    E, 


LOAD    CONDITIONS 


(C)=  -  .P0004I+  j. 001463 
(C,     4-jC2) 

=    .001464  \9I    36   18"  MHO 

SOLUTION    FOR     L 


.Es=ER(a,*ja2)+lR(coseR£jSiNeR)  (b|*ib2)*  ls=  Ir  (cosor  ±  jSiN0R)(a,^ja2)  + Er(c,  +  jC2)« 

*  i  THIS. SIGN  IS  MINUS  WHEN  THE  P.  F.  IS  LAGGING  AND  PLUS  WHEN  THE  P.  F.  IS  LEADING 


(ai+jag)    =    +  .810668  +JJ576831 


xE 


RN 


6P046 


ErnC^I^J^s)    =    ■•■    '•Se?!      +j   4613 
j/i36 


(cos  Or -j  SIN  Br)  = 


.9 

99.92 


»|p(COS0R-jSINeR)  =     4    89.93         -J43.56 

y(a|+ja2)=   +   .8  i0558'-j  .0768ni 

-    72.393      *.  ,     6.909 
<■      a347      -  ;  35.308 


Ip  (COS  Or- j  SIN  Or)    =     +  89.93-  j     43.58< 

X    (b|  Vj  bj)    =    *  91. 76  tj  235. 87 
+  8251   +  j  21212 

+  10274  -  j     3997 

|p.(coseR-jsiN0R)  (b|*-jb2)  =  +  18525  t- j  17215 

+  ERN(a|*-ja2)    =     <-  48671   +   i     4613 
^SN    ^ 


,  |p(COS0R-jSlN0f,)(a|*ja2)    =     -76240      -j:8399 


(C|+jC2)=     -.00PP4I     <-j.POI463 
X  Erm   =  60P46 


Ern(C|»JC2)    =     -  2.4( 


*■)' 


67196  *  i    21828 


Si+|p|(cosoR-jsiNOR)(a|-ia2)  =  *76.;4o     -,28.399 

Iq    =    t  73.778        *■  j  59.451 


lf(8"  96)^ +(21828)^ 
•  70  662  VOLTS  TO  NEUTRAL. 


■  V(73.778)^+  (69.4511' 


lg  =  94.75  AMPERES. 


KW,,  -  '87.1 96  X  73.778)  *  (2 1.828  X  59.45  I )  =  6.258  KW  PER  PHASE.     EFFICIENCY  =  ^  ^e"  255  ""  ^  86.33%. 


KV-A     =  (70.662  X  94.76)  =  6  694  KVA  PER  PHASE. 


PF. 


6  255  X 100 


93-42%  LEADING  . 


LOSSff  6255  -  5400  =  856  KW  PER  PHASE. 

PHASE  ANGLES —  at  full  load  the  voltage  at  the  sending  end  leads  the  voltage  at  the  receiver  end  by  the  angle 
tan    „,„°  =  tan  '.325  =  i8°oo'.  and  the  current  at  the  sending-end  leads  the  voltage  at  the  receiving -end  by  the  angle 


67196 

TAN-'  ^^-"^  '■  =  TAN- 
73.778 


,806  =  38"  52'.    HENCE  THE  CURRENT  AT  THE  SENDING-END  LEaJs  THE  VOLTAGE  AT  THE   SENDING-END  BY  THE  ANGLE  38"52' 
ANGLE  I8'00'  =  20"52'.    THE  POWER-FACTOR  AT  THE  SENDING-END  IS  THEREFORE  COS  20° 52' =  93.42%  LEADING  AT  LOAD  SPECIFIED. 


ZERO    LOAD    CONDITIONS 


EsNO"=''867l  *J46I3 


-SNO" 


'  yi4867l)2-»(46l3l^ 
=     48  889  VOLTS. 


SO  ■ 


2.462 -i  87.85 


I 

=  y  (-  2.462)^  +  87.85)^ 
|__  =     87.89  AMPERES. 


KWs 


=  (48. 57  I  X  -  2 .462)  *  (4.6  I  3  X  87. 85)  =  285 .43  KW  PER  PHASE. 


KV-ArnO"*^'^^^   X   87.89  =4  297  KVA  PER  PHASE. 

DC                    285.43  X  100  _  ^  nAf   i  ran.iof- 
Kh  SO        =    T-mI 6.64%  LEADING. 


REGULATION 

ARISE  IN  VOLTAGE  AT  THE   SENDING -END  OCCURS  OF  70  652-48  889  =  21    763  VOLTS  TO  NEUTRAL  WHEN  THE  LOAD  IS  INCREASED 
FROM  ZERO  TO  99.92  AMPERES  AT  90%  POWER  FACTOR  LAGGING  AT  THE   RECEIVER  END  WITH  CONSTANT  VOLTAGE  AT  THE  RECEIVING  END. 

PHASE   ANGLES 

AT  ZERO  LOAD  THE  VOLTAGE  AT  THE  SENDING- END  LEADS  THE  VOLTAGE  AT  THE  RECEIVER  END  BY  THE  ANGLE  TAN"'    ^g\'^|°* 
TAN"'     .0947=  5"25'    AND  THE  CURRENT  AT  THE  SUPPLY  END  LEADS  THE  VOLTAGE  AT  THE  RECEIVER  END  BY  THE  ANGLE  TAN"'  Sf^  = 
TAN  (-36.7).=  9r  36' -HENCE  THE  CURRENT  AT    THE  SUPPLY  END  LEADS  THE  VOLTAGE  AT  THE  SUPPLY  END  BY  THE  ANGLE  19  I' 36)-i5'25  I— 
86°ll'.    THE  POWER    FACTOR  AT  THE  SENDING-END  IS  THEREFORE  COS  86"  1 1'  =  6.64%  LEADING  AT  ZERO  LOAD. 


load  power- factor  is  lOO  percent,  this  rotating  triangle 
■will  equal  /  ±  j  o,  hence  it  has  no  effect  or  power  to 
rotate.  If  the  power- factor  of  the  load  is  80  percent 
the  rotating  triangle  would  have  a  numerical  value  of 
0.8  ±  j  0.(5. 

The  various  phase  angles  given  in  Chart  XIII  show 
whether  the  power-factor  at  the  supply  end  is  leading  or 
lagging.  These  various  phase  angles  are  given  to  make 
the  discussion  complete.  Actually,  in  order  to  determine 
whether  the  power-factor  at  the  supply  end  is  leading  or 
lagging,  it  is  only  necessary  to  note  if  the  supply  end 


current  vector  leads  or  lags  behind  the  supply  end  volt- 
age vector.  At  the  lower  end  of  Fig.  37  combined  cur- 
rent and  voltage  vectors  are  shown  for  this  problem, 
corresponding  to  both  load  and  zero  load  conditions. 
In  Chart  XIV  is  given  a  complete  calculation  of  the 
electrical  performance  of  problem  X,  starting  with  the 
values  for  the  auxiliary  constants  and  the  sending  end 
load  condition  known.  In  other  words  the  supply  end 
conditions  which  were  derived  by  calculation  in  Chart 
XIII  ii:i\t'  in  this  case  been  assumed  as  fixed,  and  the 
receiver   end    conditions    calculated.      The    reason    that 


CONVERGENT  SERIES  SOLUTION  FOR  LONG  LINES 


8S 


SOLUTION    FOR    Es 


Estc"  8°  0*6 


EsN~ 


y  +  18  625-' 1  * 

11  ,---t-— * 


-SN;>. 48671  VOLTS r  I 

If.. 67  186  VOLTS If 


SOLUTION    FOR    Is 


^■.. 


% 


I         8M3AMP.    ^        jBimi^ 
»o\        L  .810568-a, 


,  00 1 464=  (C)  ABSOLUTE, 
«'i.OOI4e3*C2-<^ 


d-   eoo4«X 


ls  = 


+76524  AMP. 


-j2t^399 


ls= 


-Z462  AMP 


-  j  28J99 jj^  *j  87.85  AMPERES 

* 2.482  AMPERES 


COMBINED   CURRENT   AND   VOLTAGE  VECTORS 


LOAD    CONDITIONS 


,  I  1-18.625 

(^'j.  25.  >Er^60  046  VOLTS 

>-  -j  43.56  AMPERES 


a.-  ZERO 

I     LOAD    CONDITIONS 


">«  >»  -tO 


r  =    48  889  VOLTS 
t«NO__ C3—  S'  25' 


*E:Rto*°°*«^^s 


FIG.   37 — GRAPHIC   REPRESENTATION    OF    PROBLEM    X 

Illustrating  rigorous  calculations  of  performance  when   receiving  end  conditions  are  fixed. 


there  is  a  slight  difference  between  the  receiving  end 
conditions  as  calculated  on  Chart  XIV  and  the  known 
receiving  end  conditions  is  that  the  value  for  the  sine 
in  the  rotating  triangle  (0.436)  in  chart  XIII  was  car- 
ried out  to  only  three  places,  whereas  in  Chart  XIV  it 
was  carried  out  to  four  places.  If  the  values  for  the 
rotating  triangles  had  been  carried  out  to  five  or  six 
places  in  the  calculations  in  both  charts,  the  receiving 
end  conditions  would  have  checked  exactly. 

TERMINAL  VOLTAGES  AT  ZERO  LOAD 

For  a  given  circuit  and  frequency,  the  relation  of 
the  voltage  at  the  two  ends  of  the  circuit  is  fixed.  The 
ratio  of  sending  end  to  the  receiving  end  voltage  is  ex- 
pressed by  the  constant  A.     The  ratio  of  receiving  to 

sending  end  voltage  is  expressed  by  —. .  For  example, 
problem  X,  the  sending  end  voltage  under  load  is  70  652 
volts.  If  the  load  is  thrown  off,  and  this  sending  end 
voltage  is  maintained  constant  at  70652  volts,  the  re- 

70652 

ceiving  end  voltage  will  rise  to  a  value  of    -„ = 

"  ^  0.8142 

86  775  volts  to  neutral.     The  rise  in  percent  of  sending 

100  X  86  775  —  70  652 

end   voltage   is   therefore   


22.82  percent. 


70652 


PERFORMANCE  OF  VARIOUS   CIRCUii'S 

In  Chart  XV  is  tabulated  the  complete  performance 
of  the  64  problems  for  which  the  auxiliary  constants 
are  tabulated  in  Chart  XII.  The  auxiliary  constants 
in  Chart  XII  were  applied  to  the  fixed  load  conditions 
as  stated  in  Chart  XV  for  the  receiving  end,  and  both 
load  and  zero  load  conditions  at  the  sending  end  were 
calculated  and  tabulated. 

The  object  of  calculating  and  tabulating  the  values 
for  the  64  problems  was  two  fold.  First  to  obtain  data 
on  25  and  60  cycle  problems  coverin?  a  wide  '■anee 
which  would  provide  a  basis  for  constructing  curves, 
illustrating  the  effect  that  distance  in  transmission  has 
upon  the  performance  of  circuits  and  upon  the  auxiliary 
constants  of  the  circuit.  Second,  to  give  the  student  a 
wide  range  of  problems  from  which  he  could  choose, 
and  from  which  he  could  start  with  the  tabulated  values 
as  fixed  at  either  end  and  calculate  the  conditions  at  the 
other  end.  It  is  believed  that  such  problems  will  fur- 
nish very  profitable  practice  for  the  student  and  will 
also  serve  as  a  general  guide  when  making  calculations 
on  problems  of  similar  length  and  fundamental  or  lineal 
constants.  It  is  not  intended  that  the  figures  given  for 
longer  circuits,  included  in  these  tabulations,  shall  coin- 
cide with  ordinary  conditions  encountered  in  practice. 


86 


CONVERGENT  SERIES  SOLUTION  FOR  LONG  LINES 


CHART  XIV— RIGOROUS  CALCULATION  OF  PERFORMANCE  WHEN  SENDING  END  CONDITIONS 

ARE    FIXED 


K  Ws"  1-8  '85  , 


Es=  I  22  369  VOLTS  3  PHASE  .  PFs=93  42%    LEAOINQ  . 


KV-As=20  082. 

PER  PHASE  TO  NEUTRAL 

KV-AsM=  ^2^  =  6  694.    KWs^,=  '^  =  6  265.     EsN=  ^^f^  =  70  662 .     Is°  "  '70  65  """^  ^^-'^  AMPERES. 

AUXILIARY   CONSTANTS   OF  CIRCUIT 

(  A)  =*■ -810558  + j.076831  (  B)  =  "►S  I-7486  +  j  236.868 


(a 


.8142    /  5"  24'  53' 


(bi  *  i  62) 

=  253. 083/[6814414i"  OHMS 


(C)=  -.000041  fj. 001463 
(0,     fjCj) 

=    .001464     \  9I"36'  18'  MHO 


SOLUTION    FOR  Er 


LOAD    CONDITIONS 


SOLUTION    FOR 


ER=Es(ai*ia2)-lsf'^°®^s*iSiNes)(b, +  ,b2)*  Ir=Is(°°s®s*)  siNes)(3,  +  ja,) -EgCCi^iCj)* 

*  ±   THIS  SIGN  IS  MINUS  WHEN  THE  P.  F.  IS  LAOOINO  AND  PLUS  WHEN  THE  P.  f.  IS  LEADING 


(^I*ja2)=     +  .810558  ♦  j.07683 
X  Ec,,   =     ^    '0652 


Esn(^I*J^2)    =     *■    57268      tj    5428 


(COSOs  ^jSINes)   =     +     .9342      +1,3567 


X|< 


=     4-94.75 


Ig  (coses  •^j  SIN  63)    =     +       88.62+  j     33.8 
X    (b,  +  j  bj)    =■     +       91.75  +  1235.9 


8122+      29882 
7<}73  +  j     3101 


|g(coses  +  jsiNes)(b|+,b2) 


EsN(ai*Ja2)  = 


57268 


l3(C0Ses+-jSINag)     =     +    8S.52         ,j33.8 

X    (a, +,82)    =     *     .8  I0558+,  .076831 


+    71.75!       +  ,    6.801 
-      2.597       t  ,  27.397 


|g(coses*jsiNes)(a,+ia2) 


■  j  34.198 


(C|+jC2)=    -.00004  1     ».j.00l46o 


xEc 


5428 


Esn('-1*)C2)    =      -2.897        tj    103.36 


Ig  (cosBs +j  siNeg)(a| +ja2)  =  +69.154     + 


—  |g(coses  +  jSiNes)(b|+jb2)  149 -j  23983  <- 

E^RN  =     *■      67119    -  )    18555 


-CHANGE  SIGNS  AND  ADD- 


►  -EsN(Cl-iC2)=   +    2.897 
\ff   =       72.05  1 


yi57119)^  +(13565)^ 
Erm  -  60  057  VOLTS  TO  NEUTRAL. 
K\A/     -(67  119  X  72.06  l)  + (I  8.665  X  69. 16)  =  6  399  KW  PER  PHASE, 


PF„  -  8  TJo°°  =  9a0 1%  LAGGING . 


=  Yl72.051)^    +    lf,9.16l' 
I     -  99.87  AMPERES. 


KV-A     =  '60.057  X  99.871  =  6 

"^^  "rn 


KV  A  PER  PHASE. 
LOSSm=  6  266-6  399  =  856  KW  PER  PHASE. 


EFFICIENCY 


6  399X100 


86.32%. 


PHASE  ANGLES       *t  full  load  the  voltage  at  the  receiver  end  lags  behind  the  voltage  at  the  sending-end  bythe 

ANGLETAN-I  g|-^  =  tan- '  .325  =  18' 0';  AND  THE  CURRENT  AT  THE  RECEIVER  END  LAGS  BEHIND  THE  VOLTAGe'  AT  THE  SENDING-END  BYTHE 
ANGLE  TAN-'  ^^  =  TAN-1.959  =43'50'.  HENCE  THE  CURRENT  AT  THE  RECEIVER  END  LAGS  BEHIND  THE  VOLTAGE  AT  THE  RECEIVER  END  BY  THE 
ANGLE  43- 60'  -ANGLE       1  8"0'  =  26°  50'.    THE  pOWER-FACTOR  AT  THE  RECEIVER  END  IS  THEREFORE  COS  25''50'  =  90%  LAGGING  .  


ZERO     LOAD    CONDITIONS 


Erno  -  gsN0(ai-ia2)  ^  48  898(.81056-1  .07683l)  ^  39635-)3757  _  ^^  ^^^  -  j  5667  =  60  058  VOLTS. 
"^  (a2+a2)  (.81056)^  +  1.076831)^  ■''°^'' 

iso  -^s 


(C|  a,  +  C2a2)+j  (C2  ai  -  c,  37)^ 


((-.000041  X. 8  1056 1 +  1.00 1 463  X  .07683  ll) +j  ((.00  1463  X. 8  1056) .0000  41  X.  076831)). 


(' 


I) 


|„„  =48  898  '•»-0000'92  +  i.001189L/ic  898  (.0001 1  9*  j. 00  1  7941  =  48  898  X. 00  I  798-87.92  AMPERES. 
ISO  .6629 

REGULATION 

ARISE  IN  VOLTAGE  AT  THE  SENDING-END  OCCURS  OF  70  652-48  898  =  21    754  VOLTS  TO    NEUTRAL  WHEN  THE  LOAD  IS  INCREASED 
FROM  ZERO  TO  99.87  AMPERES  AT  90.0  1%  POWER  FACTOR  LAGGING  ATI  THE  RECEIVER  END  WITH  CONSTANT  VOLTAGE  AT  THE  RECEIVING  END.' 

PHASE    ANGLES 

AT  ZERO  LOAD  THE  VOLTAGE  AT  THE  RECEIVER  END  LAGS  BEHIND  THE  VOLTAGE  AT  THE    SENDING-END  BYTHE  ANGLE 
TAN"'    ^^y  =TAN-'      .0948-5"25';  AND  THE  CURRENT  AT  THE  SENOING-END  LEADS  THE  VOLTAGE  AT  THE   SENDING-END   BY  THE  ANGLE 


CONVERGENT  SERIES  SOLUTION  FOR  LONG  LINES 
CHART  XV-CALCULATED   PERFORMANCE  OF  VARIOUS  CIRCUITS 


87 


u 

ffl 

0 
0: 

RECEIVING-END    CONDITIONS 
FIXED 

SENDING-END   CONDITIONS-CALCULATED  * 

LOAD    CONDITIONS 

LOAD    CONDITIONS 

ZERO    LOAD 

KV-A 

R 

3  PHASE 

ITO  NEUTRAL 

TO  NEUTRAL 

TO  NEUTRAL 

KV-A 

RN 

KW 

RN 

Ern 

Ir 

PFr 

% 

KV-A 

SN 

SN 

E^SN 

Is 

PFs 

% 

LINC 

DROP 

IN 

%or 

UNC 

LOSS 

IN 

%  or 

KV-A 

SNO 

KW 

SNO 

^SNO 

Iso 

PF 

SO 

% 

25        CYCLES 

/ 

2 

I300 

10  000 

« 

433.3 

346.6 
433.3 

^  77-# 

*• 

7.5- 

* 

80  LKO. 
100 

474.63 
465.09 

377.i2 
464.2/ 

6   347 
6  202 

7'4.7y 
74.99 

7*il 

94.8/ 

-9.92 
-  7.41 

8.92 
7./ 3 

/.96  3 

.5  773 

• 

.34 

• 

3 

Sooo 

20  000 

II 

/  333.3 
IkUi.k 

//  5SO 

0 

100 

IS2I.9 
1  7J6.3 

/44  9..S' 
/  783.33 

12  653 
/a   372 

I43.99t 
144.3% 

7  9.i6 
49.80 

-9.SS 
-7/2 

8.7/ 
7.00 

7.42  2 

ft  S'i9 

• 

.66 

3S00 

20  000 

//67 
II 

93  3 

1  1  (nl 

II   SSO 
n 

80  i-flG 
100 

/  278.6 
/  J53.5 

IOI7.4-S 
/2SI.22 

/2   733 
/24  (5 

loo.^l 
100.97 

7y.i8 
99.82 

-/0.J9 
-  7.49 

9.0  5 
7.2  2 

/0.85 
u 

1 

•94 

• 

7 
8 

8000 

30  000 

2667 

2/33 
2  667 

I73ZO 
It 

/54 

So  LIS. 
/oo 

2  ^agj 
2  S<«S 

2  329.8 
2  SiO.S 

/9  /2i- 
/8  i40 

IS  3.11 
/  5  3.96 

7f.« 
99.6  » 

-lOAl 
-7.6  2 

9.2  3 
7.2  6 

24. 2  f 

17313 

* 

9 

/o 

Soo'o 

So  000 

/667 

/  333 

/  667 

n  320 
II 

96.2 

80<-fl<i 
/oo 

/    8/7.3 
/   796.4 

14  59.2 
1  794.2 

19  IS4 
IS  iSS 

94.73 
96. /4 

80.2? 
99.88 

-/0.76 
-7.89 

9.47 
7.6  3 

-fO.3  2 

/7  JO-f 

m 

2.33 

It 

IZ 

20000 

* 

60  000 

1 

6667 

.S333 

i   6^7 

34  640 

/?2.5 

100 

7  303.9 
7  /y^./ 

S  S4 1.0 
7  1  81.2 

3  8  490 
37   387 

/  8  9.76 
192.37 

79.97 
4».»i^ 

-11.11 
-  7.93 

9.5  3 

7.7/ 

/44.8 

./3 

4.33 

•09 

13 
M 

22  000 
If 

83  000 
11 

7333 
II 

J'  867 
7  333 

60  8/0 
II 

0 

SOtflQ 
100 

7  742.S 
7  9/i.4 

6  4/9.6 

7  9/S.2 

Si  619 
64    S20 

137.1 
144.31 

8  2.70 

1OO.OC 

-//.4  3 
-7.89 

9.42 
7.94 

599.3 

/4-» 

50  620 
If 

//.84 

.32 

IS 
/6 

40  000 

120  000 
n 

13  333 

10  hlnl 
13  333 

i,9  Z<)o 

m.s 

80  LUC. 

100 

/4  /06 
14-  366 

II  648 
14  366 

77  147 
74  642 

IS2.iS 
I97A1 

82.ir 

100.00 

-  //.3.f 
-7.73 

9./ 9 
7.75 

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m 

J.39 

* 

69  030 

'5.66 

.3/ 

n 

IS 

2  5  000 

120  000 

8  333 

(,i,(,7 
8  333 

69  290 
If 

1Z0.3 
If 

SOLUS. 
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7  884^ 
9  03S.4 

7  ISi.l 

8  9/ao 

76   7.J4 
73  40/ 

I02.7S 
/  2  2.96 

90.74 
9S.7S1. 

-lo.n 
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7.34 
6.96 

2  IBS 

It 

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9 

68  253 

■r 

32.0/ 

.70 

11 

ZO 

40  000 

(40  000 

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10  667 
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IbS 

8oi-/?(4 
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13  270 
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//  6/0 
14412 

91    76/ 
86    86  3 

I44.SZ 
litAi 

87.44 
99.48« 

-I3.S3 

-  7.4  i 

8.84 
8.09 

2780 

n 

11.44 

79  622 

• 

3.<.92 

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21 
22 
23 
24 

2  0  000 

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6  667 

6  333 
6  667 

6,9290 
ff 

9t.2 
ff 

SOl-flG 
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7  is 2.7 

se=42.4 

7  loS,4 

75  6  82 
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75.0  « 
70  6.64 

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60  000 

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20000 

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173.7 

80/-fl«, 
100 

17  S74 
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6.55 
6.90 

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tl 

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76.69 

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25 
26 
27 
2? 

20  000 

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49.49' 
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202.04 

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29 
30 

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100 

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6.7/ 

lO.OS 

73  /40 

J9S.9 

ft 

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3.0/ 

60        CYCLES 

33 
34 

1  300 

10000 

433.3 

346.6 

433.3 

S  774 

n 

75 
If 

80  '-"^ 
100 

499.03 

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377.44 
4  64./S 

6  702 
6  2S9 

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74.94 

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98.96 

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8.90 
7J3 

4.ii8 

5  76  9 

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• 

35 
36 

s  000 

20  000 

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'333 
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nSSO 

ft 

M4.-4 

so  LUC 

loo 

/  flWi 

1  800.6 

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1  783.3 

13  333 
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7S.8-i 
99.04 

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8.70 
4.98 

18.23 
v 

//  S40 

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• 

37 
38 

3  500 

n 

30  000 

/  /67 

933 

/  /67 

II  5S0 

101 

SOLflC, 

loo 

/  34 IJ! 
1  264.0 

/  0/6.8 
/  2i/.2 

13  482 
12   537 

99..»7 
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7i.82 
98.99 

-liJ3 
-8-Si 

Sit 

7.2  2 

2S.9  3 

II   527 

2.2s 

Z9 

40 

S  000 

1' 

30  000 

2  667 

^  /33 
2  667 

17320 

IS4 
II 

8oi.fl<; 
100 

3   0  73.« 

2  8 '-'.7 

2  327.9 
2  864./ 

20  268 
/«  830 

ISI.CS 
IS3.73 

7S.74 
9  8.94 

-(7.07 
-8.72 

9-/ 3 
7.39 

58.4  3 

'7286 

338 

4' 
42 

S  000 
f 

so  000 

/  667 
If 

/  333 
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n  320 
fi 

9i>.Z 

f 

so  LHi 

loo 

/   g79.? 

/  806./ 

/  4  56.2 
/  794./ 

20  33; 
/8  845 

92.43 
9S.84 

77..90 
99.3  3 

-/7.3! 
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9.24 
7.6  2 

96.29 

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5.59 

V 

•2  2 

43 

44 

zoooo 

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6  667 

5333 
6  66  7 

34  (,iiO 

9 

801.1s. 
loo 

7.59  7.8 
/  ?  4  3.0 

i  830./ 
7  /  80.2 

40  97i 
37  773 

I8S/I2 
I9I.7S 

74.73 
99./ 3 

-18.29 
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9.32 
7.70 

3  5  7.9 

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0 

34  4  50 

Jr 

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■21 

0 

4i 
46 

2  2  000 

88  000 

7  333 

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7333 

SO  8IO 

144.4 
If 

SOLfS. 

100 

7  5  7S.7 
7  t  l^.i 

6  380/) 

7  9/S.3 

5  9  92,f 
54    869 

I2t/n 
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g4J8 
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-17.94 
-7.99 

8.74 
7.94 

/  409 

If 

-♦9  7/0 

2  8.35 

m 

.6/ 

47 
48 

-«4  000 

120  000 

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It 

lOCil 
1333-i 

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If 

I92.S 

so  Lac, 
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13  796 

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II  S79 
14  3US 

81  7/0 
74   735^ 

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192.22 

83.93 

100.00 

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-7.86 

8-55 

7.74 

2  528 

14^9 

ff 

6  7  900 

37.28 

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49 

2.S  000 

120  000 

If 

8  333 
II 

6  <S67 
8  333 

69290 

So  Lite. 
100 

7  OSJ.; 
9  4  73.0 

7  07i./ 

8  9-<9.6 

7  9  000 
70599 

S9.ti 

I3  4.IB 

99.g? 
94.47- 

-'4.01 
-1.89 

iJ2 
7.40 

-»  7S9 

ff 

75.47 

m 

63  357 

r 

7i." 

• 

'•59 

5' 
S2 

40  000 

/fO  00c 

13  333 

10  667 
/3  333 

80S3O 

IbS 

So  LKC 

/oo 

//  827 
/4  6  66 

//  46/ 
/4  438 

96  727 
84  862 

'22.27 
172.82 

96.fO 
9»'.44 

-I9.i1 
-4.99 

7<44 
t.29 

i  060 

89.78 

m 

73  938 

8 '.96 

'.4  8 

S3 
St 

20000 

120  oco 

6  667 
If 

5  333 

6  667 

(,9390 

9 

96.2 

SOiflG. 

/oo 

6  972? 
9  06/.7 

s6Z6ii 
7  231it 

72  747 
6-3  8/0 

9XSS 
142.01 

80.^^ 
79.89» 

-4.99 
+  7.91 

S.SO 
8-S9 

6  523 

• 

208t 

56  101 

m 

/'4.27 

3.20 

W 

6S 
Si. 

Co  000 

200  000 
II 

ZOOOO 

16  000 
20000 

IIS  SOO 

;73.3 

ZOLte. 
/oo 

IS  72S 
2  4  796 

l<f   90s 
21  658 

/26.S4/ 
(09/89 

/-I8.00 
227.09 

90  J  8. 
87.3'C 

-f.Si 

♦  5-47 

5.68 
8:29 

/6  330 

-»76.4 

• 

9J636 

•■  '■*.'< 

2.92 

S7 
SS 

20  000 

MO  COO 

fi 

3  333 
6  667 

80  830 

82.5 

/oo 

/o  089 
//  OI4 

S79CJI 
7  53?.6 

74  /82 
64  377 

llt.OI 

1 7/.0S 

5  7.4ii 
«8.4  j 

+  8.2  2 
+  20.3  J 

8.69 
l\09 

8626 

5-.i.8 

54  4  9-» 

/.SS.3 

6.33 

*    * 

SI 
60 

SO  000 

200  COO 

If 

It,  667 

/3  333 
/6  667 

IIS  SOO 

144.* 

8oi.«<:. 
loo 

21  139 
2  3  944, 

14  343 
(8  7^-7 

113  606 
96987 

ist.o? 

24i.i4 

6  7.85 
7?.  3  J 

+  A64 
+  /6.0  3 

7.St 

I2.S4 

17  217 

/  Oi7 

77  93« 

J  2  0,* 

6.'4 

6/ 
62 

16  000 

MO  000 
n 

^  000 
*   - 

4  000 
.5  000 

80  830 

dl.Si 

m 

SOLm 
100 

10  233 
9  9 /ft* 

-f  ffO/J! 
6  2  30.f 

59046 
SI  327 

/7  3.30 
'»3.2.» 

4t.9> 
6  2.8/' 

+  2C9i 
+  36J0 

20j3-i 
2*.lO 

7  690 

998.8 

.»  /  2  /  3 

'2.9< 

63 
64 

40  000 
/I 

20000c 

'3  333 

10  U7 

/3  333 

f  l£  SCO 
tt 

11S.S 

SOLUS. 

/oo 

J/f-»8 
2i  7 SO 

13  24* 
/«  05O 

93  72J 
80  /06 

23*. '7 
J7/.J2 

5i.8d 
7  3.fc6 

+  /».*■' 
+  30.*t 

I4.t2 

iojS 

IS  223 

/  907 

5  9  07.# 

* 

2i7.7 

'2.  J  J 

The  above  performances  are  based  upon  values  for  the  auxiliary  constant's  as  Riven  on  Chart  XII 


CHAPTER  X 
HYPERBOLIC   FUNCTIONS 


In  the  consideration  of  the  hyperbolic  theory  as  applied  to  transmission  circuits,  the  writer  desires  to 
express  his  high  appreciation  of  the  excellent  literature  already  existing.  Dr.  A.  E.  Kennelly's  pioneer 
work  and  advocacy  of  the  application  of  hyperbolic  functions  to  the  solution  of  transmission  circuits  has 
been  too  extensive  and  well  known  to  warrant  a  complete  list  of  his  contributions.  His  most  important 
treatises  are  "Hyperbolic  Functions  Applied  to  Electrical  Engineering",  igi6;  "Tables  of  Complex  Hyper- 
bolic and  Circular  Functions",  1914;  "Chart  Atlas  of  Hyperbolic  Functions",  1914,  which  provides  a  ready 
means  of  obtaining  values  for  complex  functions,  thus  materially  shortening  and  simplifying  calculations, 
and  "Artificial  Electric  Lines",  1917. 

"Electrical  Phenomena  in  Parallel  Conductors"  by  Dr.  Frederick  Eugene  Pernot,  1918,  is  an  excellent 
treatise  on  the  subject  and  contains  valuable  tables  of  logarithms  of  real  hyperbolic  functions  from  x  :=  0 
to  X  =:  2.00  in  steps  of  o.ooi. 

An  article  "Long-Line  Phenomena  and  Vector  Locus  Diagrams"  in  the  Electrical  World  of  Feb.  I, 
1919,  p.  212,  by  Prof.  Edy  Velander  is  an  excellent  and  valuable  contribution  on  the  subject,  because  of  its 
simplicity  in  explaining  complicated  phenomena. 

To  employ  hyperbolic  functions  successfully  in  *ht  solution  of  transmission  circuits  it  is  not  necessary 
for  the  worker  to  have  a  thorough  understanding  of  how  they  have  been  derived.  On  the  other  hand  it 
is  quite  desirable  to  understand  the  basis  upon  which  they  have  been  computed.  A  brief  review  of  hyper- 
bolic trigonometry  is  therefore  given  before  taking  up  the  solution  of  circuits. 


CIRCULAR  angles  derive  their  name  from  the  fact 
that  they  are  functions  of  the  circle,  whose  equa- 
tion is  x"  -\-  y"  =^  I.  Tabulated  values  of  such 
functions  are  based  upon  a  radius  of  unit  length.  The 
geometrical  construction  illustrating  three  of  the  func- 
tions, the  sine,  cosine  and  fangent  of  circular  angles  is 
indicated  in  Fig.  38.  The  angle  AOP,  indicated  by  full 
lines  in  the  positive  or  counter-clockwise  direction,  has 
been  drawn  to  correspond  to  one  radian.  The  radian 
is  an  angular  unit  of  such  magnitude  that  the  length  of 
the  arc  which  subtends  the  radian  is  numerically  equal 
tc  that  of  the  radius  of  the  circle.  Thus,  the  number 
of  radians  in  a  complete  circle  is  2  t.  Expressed  in 
degrees  the  radian  is  equal  approximately  to  57°  17' 
44.8".  The  segment  AOP  of  any  angle  AOP  of  one 
radian  has  an  area  equal  to  one-half  the  area  of  a  unit 
square.  Therefore  the  angle  may  be  expressed  in 
radians  as, — 

Length  of  arc  s  X  area 

(radius)' 


or 


Circular  angle  =  ■ 


radians 


radius 

Circular  functions  are  obtained  as  follows,- 
S  X  area 
(radius)' 
Y_ 
R 

R 
Tangent  0  =  -y- 


Sine  0         =  "5" 


Cosine  $     =  -5" 


The  variations  in  the  circular  functions,  sine, 
cosine  and  tangent  are  indicated  graphically  in  Fig.  39 
for  a  complete  revolution  of  360  degrees.  Since  for  the 
second  and  each  succeeding  revolution  these  graphs 
would  simply  be  repeated,  circular  functions  are  said  to 
have  a  period  equal  to  2  t  radians.  In  other  words, 
adding  ^  t  to  a  circular  angle  expressed  in  radians  does 
not  change  the  value  of  a  circular  function. 


REAL  HYPERBOLIC  ANGLES 

Real  hyperbolic  angles  derive  their  name  because 
they  are  functions  of  an  equilateral  hyperbola.  A 
hyperbola  is  a  plane  curve,  such  that  the  difference  be- 
tween the  distances  from  any  point  on  the  curve  to 
two  fixed  points  called  the  foci  is  constant.  In  an  equi- 
lateral hyperbola.  Fig.  40,  the  asymptotes  OS  and  OS' 
are  straight  lines  at  right  angles  to  each  other  and  make 
equal  angles  with  the  X-axis.  The  hyperbola  continu- 
ally approaches  the  asymptotes,  and  meets  them  at 
infinity.     The  equation  of  such  a  hyperbola  is  x"  —  y* 

The  hyperbolic  angle  AOP  of  Fig.  40,  called  for 
convenience  **,  has  been  drawn  so  as  to  correspond  to 
an  angle  of  one  hyperbolic  radian,  or  one  "hyp"  as  it  is 
usually  designated.  Hyperbolic  angles  are  determined 
by  the  area  of  the  sector  they  enclose.  Thtis  the  hyper- 
bolic angle  of  one  hyp  AOP,  encloses  an  area  AOP  of 
one-half,  or  the  same  as  the  area  AOP  of  the  corre- 
sponding circular  angle  of  Fig.  38.  It  should  be  ob- 
served here  that  although  one  circular  radian  subtends 
an  angle  AOP  of  57°  17*  44-8",  one  hyperbolic  radian 
subtends  a  circular  angle  AOP  of  37°  17'  33.67" 
(0.65087  circular  radian). 

In  the  same  way  as  for  the  circle  the  hyperbolic 
angle  may  be  expressed  in  radians  as, — 

Length  of  arc  2  X  area 

p  "'^     (radius)' 

where  p  =  the  integrated  mean  radius  from  0  to  AP. 
As  an  illustration,  the  length  of  the  arc  AP,  Fig,  4Q 


*A  "hyperbolic  angle",  in  the  sense  above  described,  is  not 
the  opening  between  two  lines  intersecting  in  a  plane,  but  a 
quantity  otherwise  analogous  to  a  circular  angle  and  the  argu- 
ment x  of  the  function  sinh  x,  cosh  x,  tanh  x,  etc.  The  use  of 
the  term  hyperbolic  angle  can  only  be  justified  by  its  con- 
venience of  anology. 


HYPERBOLIC  FUNCTIONS 


is  1.3167  and  the  mean  integrated  radius  to  arc  AP  is  variations   in   hyperbolic   functions  are   indicated 

A-3I&7-  aphically  in  Fig.  41  for  hyperbolic  angles  up  to  ap- 

Hyperbolic  functions,  distinguished  from  circular  proximately  2.0  hyps  for  the  sine  and  cosine  and  up  to 

lunctions  by  the  letter  h  affixed,  are  obtained  as  fol-  3.0  hyps  for  the  tangent, 

lows: —  Hyperbolic  functions  have  no  true  period,  but  add- 

y      3Vf 


+  YAXIS 


YAXIS 


XAXI3 


NOTE 
ANULb  AOP  IS   DRAWN 
CORRE8PONDINQ  TO  ONE       y 
HYPERBOLIC  RADIAN  OR 
-HYP" 


FIG.   38 — REAL   CIRCULAR   ANCLES 
X'  +¥'  =  J 


FIG.   40 — REAL   HYPERBOUC  ANCLES 

x*  — r  =  / 


1^1  rP 


2  7T  RADIANS 
4  QUADRANTS 


360* 


ONE  CIRCULAR  RADIAN  -<  ' 
FIG.   39 — GRAPHS  OF  CIRCULAR  FUNCTIONS 


6.2831863072 
=  206264.8062' 
■en  7' 44.8" 
■  Br.28678 


4.0 
3.6 

2JS 

c 

f 

62.0 
u. 

4 

7/ 

^^ 

<. 

^ 

^ 

-fSSc 

b?rff 

mTh) 

/ 

0 

5    '1. 

6       1 

6       2 

.0  ,4 

6     4. 

0 

Hyperbolic  angle  6  ■■ 


Cosh  e 


Length  of  arc  AP 
Length  of  mean  radius 
X 


radians. 


Sink   0  = 

Tank  e  = 


OA 

y 

OA 
Y 


HYPERBOLIC  RADIANS  I  HYPSl 

KIG,    41 — GRAPHS     OF     HYPERBOUC 
FUNCTIONS 

ing  a  i"  »  ;'  to  the  hyperbolic  angle  does  not  change  the 
values  of  the  functions,  hence  these  functions  have  an 
imaginary  period  oi  2  ^  j. 

Circular  functions  can  be  used  to  express  the  phase 
relations  of  current  and  voltage,  but  not  the  magnitude, 
or  size,  whereas  hyperbolic  functions,  continually  in- 


90 


HYPERBOLIC  FUNCTIONS 


creasing  or  decreasing,  can  be  used  to  express  the  mag- 
nitude of  current  in  a  long  circuit. 

In  Fig.  42  is  shown  a  circular  angle  corresponding 
to  one  circular  radian  divided  into  five  equal  parts,  each 
of  0.2  radian.  Assuming  unity  radius,  each  of  the  arcs 
will  have  a  constant  length  of  0.2  and  a  constant  mean 
radius  of  i.e.  In  Fig.  42  is  shown  a  hyperbolic  angle 
corresponding  to  one  hyperbolic  radian  divided  into  five 
equal  hyperbolic  angles  each  of  0.2  hyperbolic  radian. 
In  this  case  the  length  of  the  arcs  corresponding  to  each 
subdivision  increases  as  the  hyperbolic  angle  increases. 
The  lengths  of  the  corresponding  integrated  mean  radii 
vectors  also  increase  with  the  angle.  By  dividing  the 
length  of  the  arc  of  any  of  the  five  subdivisions  by  the 
length  of  the  mean  radius  for  that  subdivision  it  will  be 
seen  that  each  subdivision  represents  0.2  hyps. 

From  the  above  it  will  be  evident  that  in  radian 
measure,   the  magnitudes   of   circular   and  hyperbolic 


'^>-. 


1.000 

CIRCULAR  RADIAN 


000--. VA 

HYPERBOLIC  RADIAN 


nc. 


42 — SUBDnnSIOK  OF  A  CIRCULAR  AND  A  HYPERBOLIC    RADIAN    INTO    FIVE    SECTORS    OF    O.i 

RADIAN  EACH 


angles  are  similarly  defined  with  reference  to  the  area 
of  circular  and  hyperbolic  sectors. 

COMPLEX  ANGLES  AND  THEIR  FUNCTIONS 

A  complex  angle  is  one  which  is  associated  with 
both  a  hyperbolic  and  a  circular  sector.  If  the  complex 
angle  is  hyperibolic,  its  real  part  relates  to  a  hyperbolic 
and  its  imaginary  to  a  circular  sector.  On  the  other 
hand,  if  the  complex  angle  is  circular,  its  real  part  re- 
lates to  a  circular  and  its  imaginary  part  to  a  hyper- 
bolic sector.  Complex  hyperbolic  trigonometry  and 
complex  circular  trigonometry  thus  unite  in  a  common 
geometrical  relationship. 

In  the  following  treatment  for  the  solution  of  trans- 
mission circuits  by  hyperibolic  functions,  only  hyperbolic 
complex  angles  will  enter  into  the  solution.  Such  a 
conTplex  angle  will  then  consist  of  a  combination  of  a 
"real"  hyperbolic  sector  and  a  so-called  "imaginary"  or 
circular  sector.  The  circular  sector  will  occupy  a  plane 
inclined  at  an  angle  to  the  plane  of  the  hyperbolic  sector. 
In  other  words,  the  complex  angle  will  be  of  the  three- 
dimensional  order.  The  construction  of  such  a  complex 
angle  may  be  difficult  to  follow  if  viewed  only  from  one 
direction.     In  order  to  illustrate  the  form  that  a  com- 


plex angle  takes,  the  construction  for  the  cosine  of  a 
hyperbolic  complex  angle  is  illustrated  by  Fig.  43. 

CONSTRUCTION   FOR  COSH   6 

The  construction,  Fig.  43,  assumes  that  the  real 
part,  that  is  the  hyperbolic  sector  subtends  an  angle  of 
one  hyperbolic  radian  and  the  imaginary  part,  that  is  the 
circular  sector,  subtends  an  angle  of  one  circular  radian. 
This  hyperbolic  complex  angle  has  therefore  a  numerical 
value  oi  I  -\-  j  I  hyperbolic  radian.  These  numerical 
values  embrace  sectors  sufficiently  large  for  the  pur- 
pose of  clear  illustration.  The  actual  construction  for 
obtaining  the  complex  function  cosh  (^i  -f  /  6^)  =  cosh 
(i  -{•  f  I  hyperbolic  radians)  may  be  carried  out  a.1  fol- 
lows:— 

On  a  piece  of  stiff  card  board  lay  out  to  a  suitable 
scale  the  hyperbolic  sector  d^  =  EOC,  equal  to  one  hyp 
as  shown  in  the  upper  left  hand  corner  of  Fig.  43.  This 
may  readily  be  plotted  by  the  aid 
of  a  table  of  real  hyperbolic  func- 
tions for  say  each  one  tenth  of  a 
hyp  up  to  and  including  one  hyp. 
These  are  then  plotted  on  the 
cardboard  and  joined  with  a 
curved  line  thus  forming  the  arc 
EC  of  Fig.  43.  The  ends  of  the 
arc  are  then  joined  with  O  by 
straight  lines.  The  real  part  of 
this  hyperbolic  complex  angle  is 
then  cut  out  of  the  cardboard. 

The  circular  part  ;  6^  of  this 
complex  angle  is  traced  upon  the 
cardlxjard  as  follows  :  — •  With 
radius  equal  to  cosh  0^  (to  the 
same  scale  as  used  when  trac- 
ing the  hyperbolic  sector  0  J  draw  the  arc  DOF  of  a 
length  such  that  the  angle  DOF  is  57°  17'  44.8"  (on: 
circular  radian).  Join  the  ends  of  the  arc  to  0  witi 
straight  lines.  The  circular  part  j  6  ^  oi  this  complex 
angle  is  now  cut  out  of  the  piece  of  cardboard.  This 
gives  models  of  the  two  parts  of  the  complex  angle 
which  may  be  arranged  to  form  the  complex  angle 
I  -{-  j  I  hyps.  These  two  models  are  shown  at  the  top 
of  Fig.  43. 

The  two  parts  of  the  complex  angle  are  arranged 
as  follows : — -Upon  a  drawing  board  or  any  flat  surface 
occupying  a  horizontal  plane,  place  the  hyperbolic  sec- 
tor S,  in  a  vertical  position.  The  plane  of  this 
hyper])olic  sector  will  then  be  at  right  angles  to  the 
plane  of  the  drawing  board.  The  circular  sector  ;  B^ 
is  now  placed  in  a  vertical  position  just  back  of  the 
hyperbolic  sector.  The  toes  0  of  each  sector  will  then 
coincide,  as  well  as  the  line  OD  of  the  circular  sector 
with  the  line  OC  of  the  hyperbolic  sector.  The  top  of 
the  circular  sector  is  now  turned  back  so  that  the  plane 
of  the  circular  sector  lies  at  an  angle  with  the  vertical 
plane  occupied  by  the  hyperbolic  sector.  This  displace- 
ment angle  between  the  planes  of  the  two  sectors  is 


HYPERBOLIC  FUNCTIONS 


QI 


k COSHO,-  1.643081 V 

HVPERBOLIO  8E0T0R 


RADIUS-CCSHe, -1.643081 -■ 
CIRCULAR  SECTOR 


^ 


QUDERMANNIAN    COMPLEMENT 
OF  THE  COMPLEX  HYPERBOLIC 
ANGLE  OF  THE  CIRCUIT -Bq 
^08  80-55^ -SECHe, 


FRONT  ELEVATION 


LEFT  END  ELEVATION 


_COSH^j^008_92__J+x  axis) 

+  6.8337~  VECTOR  OF  REFERENCE 


TOP  PLAN  OF  BOARD  (MODEL  REMOVED) 
AHOWINO  0O8H  (0,  ♦{  Og  )  TRACED  UPON  BOARD 


REAR  ELEVATION 


MATHEMATICAL  SOLUTION 
COSH  (e,  +  j  8]  )  -  (COSH  8,  COS  82  +  jSINH  8,  SIN  Sg) 

LOa  COSH  8, -0.1  88,389 

LOG  COS  82-T-732.639 

7.821.026 


LOG  SINH  8,-0.070.1 1 2 
LOG  SIN  82  -  T  .926,039 


1.996,161. 
COSH  (81  + j  82) -0.8337 +j  0.9889 
-1, 297/49*  62' 06' 
43 — GRAPHICAL  CONSTRUCTION  FOE  THE  HYPERBOLIC   COSINE   OF   THE   COMPLEX   ANGLE 

«!  +  /*i  =  1+  ii  Hyperbolic  Radians. 


circular  sector  of  this  complex 
angle  is  moved  in  the  forward 
direction  through  an  angle  of 
49°  36'  18"  so  that  the  plane  of 
the  circular  sector  assumes  an 
angle  of  90°  oc/  00" — 49°  36^ 
18"  =  40°  23'  42"  with  the  hor- 
izontal plane  of  the  drawing 
board.  From  the  end  of  the  cir- 
cular sector  (point  F)  thus  in- 
clined, a  plummet  may  tie  sus- 
pended until  it  meets  the  hori- 
zontal plane  of  the  drawing  board 
at  the  point  /  of  the  illustration. 
In  other  words, the  point  F  is  pro- 
jected orthogonally  onto  the  hori- 
zontal plane  of  the  drawing  board. 
A  top  view  of  the  drawing 
board,  with  the  model  removed, 
is  illustrated  in  the  lower  left 
hand  corner  of  Fig.  43.  The  line 
OF  {1.297  /49°  52'  05'')  traced 
upon  the  horizontal  drawing 
board,  is  a  vector  representing 
the  complex  cosine  of  the  com- 
plex angle  0^  -{-  j  6^  —  i  -\-  j  i 
hyperbolic  radians.  This  com- 
plex cosine  has  rectangular  co- 
ordinates of  -f  0.8337  and  -1- 
;■  0.9889. 

At  the  bottom  of  Fig.  43  is 
given  the  mathematical  expres- 
sion for  the  exact  solution  for 
the  cosine  of  a  complex  hyper- 
bolic angle  following  the  con- 
struction illustrated.  There  are 
numerous  other  mathematical 
equations  with  their  equivalent 
geometrical  constructions  which 
will  produce  the  same  values  for 
the  cosine,  but  the  above  is  prob- 
ably as  easy  to  follow  as  any,  and 
will  therefore  be  used  exclusively 
hereafter. 


known  as  the  "gudermannian  complement"  of  the  hy- 
perbolic angle  d.  It  will  be  referred  to  as  6o-  The 
front  elevation  of  Fig.  43  illustrates  how  these  two  sec- 
tors would  appear  when  viewed  from  the  front.  To 
the  right  of  this  illustration  is  shown  how  these  two 
sectors  would  appear  when  viewed  from  the  left  hand 
end  of  tlie  model.  The  displacement  angle  Ba  has  a 
value  for  this  particular  complex  angle  of  49°  36'  18". 
This  numerical  value  is  determined  by  virtue  of  the 
fact    that    this    displacement    angle    has    a    cosine    of 

•r  T 

=  0.64805  or  cosine  of  6^  =  sech  6^ 


cosh  611        1.543081 

:=  0.64805.    It  has  a  sine  of  tanh  6,  =  0.76159. 

The  angle  whose  cosine  is  0.64805  and  whose  sine 
is  0.76159  is  49°  36'  18".     Thus  the  top  part  of  the 


CONSTRUCTION    FOR    SINH    6 

The  construction  for  the  sine  of  the  complex  hy- 
perbolic angle  /  +  ;  /  is  indicated  in  Fig.  44.  In  this 
case  the  same  construction  may  be  used  for  obtaining 
the  sinh  as  for  determining  the  cosh  of  the  complex 
angle  with  the  following  two  exceptions. 

The  circular  sector  is  made  one  quadrant  (90°) 
larger.  In  other  words  the  angle  DOF'  is  90°  4-  57° 
17'  44.8"  or  147°  17'  44.8"  as  indicated  by  Fig.  44.  It 
occupies  the  same  plane  as  when  determining  the  cosh 
of  the  angle  but  is  simply  extended  in  the  forward  di- 
rection through  one  quadrant,  as  indicated  by  the  dotted 
lines  of  Fig.  44.  The  plummet  is  again  suspended,  this 
time  from  point  F'  upon  the  horizontal  board,  which  it 


HYPERBOLIC  FUNCTIONS 


meets  at  point  /'.  The  other  difference  is  that  the  sine 
OF'  is  read  off  from  the  Y  axis  as  the  vector  of  refer- 
ence in  place  of  the  X  axis  as  in  the  case  of  the  cosine. 
Thus  the  circular  sector  has  been  carried  forward 
through  an  angle  of  90  degrees  in  the  circular  angle 
plane  and  the  vector  of  reference  has  been  advanced  90 
degrees  in  the  horizontal  plane  of  reference.  The  sine 
of  this  angle  is  1.446  /63°  56'  3/^  and  has  rectangular 
components  of   0.6349  -f-  ji.sgSs.     The  mathematical 


«c^^ 


\ 


■■*r 


o^ 


FRONT  '  ELEVATION 


LOG  SINH  01=0.070  112 

LOO  COS  02 -7.732  638 

7,802748 


SINH  (91+ j  02)- 0.6349  +1  1.2985 
-  I  446  /e3'  66'  37* 

NOTE -THE  CONSTRUOTJON  FOR  THE  COSINE  OF  THE  COMPLEX 
HYPERBOLIC  ANGLE.  MAY  ALSO  BE  USED  FOR  DETERMINING  THE 
SINE  OF  THE  ANGLE  WITH  THE  FOLLOWING  CHANGES:  -  THE  CIR- 

~XA-XiS _t:^+x  aJ(IS   CULAR  SECTOR  MUST  BE  EXTENDED  THRU  ONE  QUADRANT  AND 

THE  SINE  MEASURED  FROM  THE  Y  AXIS  AS  THE  VECTOR  OF  REFER- 
TOP  PLAN  OF  BOARD  (MODEL  REMOVED)  ENCE  IN  PLACE  OFT  HE  X  AXIS  AS  IN  THE  CASE  OF  THE  COSINE. 

SHOWING  81NH  "(©i  +  jej)  TRACED  UPON  BOARD    C 

FIG.   44 — GRAPHICAL    CONSTRUCTION    FOR   THE    HYPERBOLIC  SINE  OF  THE  COMPLEX  ANGLE 

6i  +  jffi  =^  J  -^  ji  hyperbolic  radians. 


expression  for  exact  solution  for  the  sine  of  a  complex 
angle  likewise  accompanies  the  illustrated  geometrical 
construction. 

MODEL    FOR    ILLUSTRATING   THE    FUNCTIONS    OF    A 
COMPLEX   ANGLE 

Dr.  Kennelly  has  recently  constructed  a  model*  for 
illustrating  complex  angles  and  for  obtaining  approxi- 
mate values  for  the  functions  of  such  angles.  Drawings 
made  from  photographs  of  this  model  are  shown  in  Figs. 
45,  46  and  47.  The  construction  of  a  complex  angle  as 
above  described  is  that  employed  by  Dr.  Kennelly  in 
building  his  model.  Since  the  model  is  applicable  to 
tracing  out  numerous  complex  angles,  it  may  seem  a  little 
difficult  at  the  start.  It  was  therefore  thought  desirable 
to  precede  the  description  of  the  model  which  is  appli- 
cable to  the  solution  of  so  many  angles  with  a  similar 
solution  of  a  single  definite  complex  angle.  With  the 
procedure  for  the  solution,  as  given  above,  for  cosh  and 
sinh  of  I  -j-  y  ^  hyperbolic  radians  in  mind,  it  is  believed 

*This  model  was  described  in  a  paper  read  by  him  at  a 
meeting  of  the  American  Academy  of  Arts  and  Sciences  in 
April  1910. 


that  Dr.  Kennelly's  description  of  the  model  and  its  appli- 
cation in  determining  the  cosh  and  sinh  of  complex  angles 
may  be  followed  as  given  in  the  following  paragraphs. 

DESCRIPTION  OF  MODEL 
In  this  model,  the  cosine  or  sine  of  a  complex  angle,  either 
hyperbolic  or  circular,  can  be  produced,  by  two  successive 
orthogonal  projections  onto  the  XY  plane,  one  projection  being 
made  from  a  rectangular  hyperbola,  and  the  other  projection 
being  then  made  from  a  particular  circle  definitely  selected 
from  among  a  theoretically  infinite  number  of  such  circles,  all 
concentric  at  the  origin  O,  which  circles,  however,  are  not 
P  coplanar.      The    selection    of 

the  particular  circle  is  deter- 
mined by  the  foot  of  the  pro- 
jection from  the  hyperbola. 
This  effects  a  geometrical  pro- 
cess which  is  easily  appre- 
hended and  visualized ;  so  that 
once  it  has  been  realized  by 
the  student,  the  three-dimen- 
sional artifice  is  rendered  su- 
perfluous, and  he  can  roughly 
trace  out  a  complex  sine  or 
cosine  on  an  imaginary  draw- 
ing board,  with  his  eyes  closed. 
The  model,  however,  possesses 
certain  interesting  geometrical 
properties  as  a  three-dimen- 
sional structure. 

A  drawing  made  from  a 
photograph  of  the  model  is 
shown  in  Fig.  45.  On  an  ordi- 
nary horizontal  drawing  board 
53-5  by  31.8  cm.,  is  a  horizon- 
tal rod  AB,  which  merely 
serves  to  support  the  various 
brass-wire  semicircles,  and  a 
semihyperbola,  in  their  proper 
positions.  The  axis  of  AB  in 
the  XY  plane,  on  the  upper 
surface  of  the  board,  is  a  line 
of  symmetry  for  the  structure, 
which,  if  completed,  would  be 
formed  by  full  circles  and  a 
complete  hyperbola.  For  con- 
venience, however,  only  the 
half  of  the  structure  above  the 
XY  plane  is  presented,  the 
omission  of  the  lower  half 
being  readily  compensated  for 
in  the  imagination. 
The   eight  wire   semicircles  are   formea  with  the   following 

2.0,  1.020. ...  1.081...,  1.185.. 


MATHEMATICAL  SOLUTION. 
SINH  (8|  +  i  92)=lSINHe,'COS02+jCOSHe.81N  Oj) 


LOG  OOSH0|  — 0.188  389 

LOO  SIN  02  -7.825  039 

0.113428 


respective  radii,  in  decimeters: 

1.337.  ••.  I-S43---.  l.Slo...,  ana  2.150...,  wmcn  are  me  re- 
spective cosines  of  o,  0.2,  0.4,  0.6,  0.8,  l.o,  1.2,  and  1.4  hyperbolic 
radians,  according  to  ordinary  tables  of  real  hyperbolic  func- 
tions. These  successive  semi-circles  therefore  have  radii  equal 
to  the  cosines  of  successively  increasing  real  hyperbolic  angles 
01,  by  steps  of  0.2,  from  o  to  1.4  hyperbolic  radians,  inclusive. 
All  of  these  semicircles  have  their  common  center  at  the  origin 
O,  in  the  plane  X  0  Y,  of  the  drawing  board.  The  planes  of 
the  semicircles  are,  however,  displaced.  The  smallest  circle  of 
unit  radius   (l  decimeter),  occupies  the  vertical  plane  X  O  Z, 


XAXIS 


FIG.   45 — DRAWING   FROM    A    PHOTOGRAPH    OF   A   GEOMETRICAL   MODEL 

For  the  orthogonal  projection  of  the  sines  and  cosines  of 
complex  angles.     This  model  was  developed  by  A.  E.  Kennelly. 


HYPERBOLIC  FUNCTIONS 


93 


in  which  also  lies  the  rectangular  semi-hyperbda  X  O  H. 

Angular  distances  corresponding  to  0.2,  0.4 14  hyperbolic 

radians,  are  marked  off  along  this  hyperbola  at  successive 
corresponding  intervals  of  0.2.  The  cosines  of  these  angles, 
as  obtainable  projectively  on  the  O  X  axis  are  marked  off 
between  C  and  B  along  the  brass  supporting  bar,  and  at  each 
mark,  a  semicircle  rises  from  the  X  Y  plane,  at  a  certain  angle 
e„  with  the  vertical  X  0  Z  plane.  This  displacement  angle  i« 
determined  by  the  relation, — 
X 


cus  $a  ■ 


■  ^  seek  01 


cosh  81 

Where  ffi  is  the  partictilar  hyperbolic  angle  selected.  This 
means,  as  is  well  known,  that  the  displacement  angle  6a  between 
the  plane  of  any  semicircle  a,nd  the  vertical  plane  Z  O  X  'n 
equal  to  the  gudermannian  of  the  hyperbolic  angle  ft. 

The  model  is,  of  course,  only  a  skeleton  structure  of  eight 
stages.  If  it  could  be  completely  developed,  the  number  of 
semicircles  would  become  infinite,  and  they  would  form  a 
smooth  continuous  surface  in  three  dimensions.  Along  the 
midplane  Z  O  Y,  all  01  these  circles  would  have  the  same  level, 
raised  one  decimeter  above  the  horizontal  drawing  board  plane 
of  reference  X  0  Y.  The  circles  would  increase  in  radius 
without  limit,  and  would  cover  the  entire  X  O  Y  plane  to 
infinity,  the  hyperbola  extending  likewise  to  infinity  toward? 
its  asymptote  O  S,  in  the  X  O  Z  plane.  The  actual  model  is 
thus  the  skeleton  of  the  upper  central  sheet  of  the  entire 
theoretical  surface,  near  the  origin. 

The  semicircles  are  also  marked  off  in  uniform  steps  jf 
circular  angle.  Each  step  is  taken,  for  convenience,  as  nine 
degrees,  or  one  tenth  of  a  quadrant.  Corresponding  angiilar 
steps  on  all  of  the  eight  semicircles  are  connected  by  thin  wires, 
as  shown  in  the  illustrations. 

A  front  elevation  of  the  model,  taken  from  a  point  on  the 
O  Y  axis — 15  units  from  O,  is  given  in  Fig.  46.  It  will  be  seen 
that  any  tie  wire,  connecting  corresponding  circular  angular 


KG.   46 — FRONT  ELEVATION  OF  MODEL 

From  a  point  on  the  O  7  axis,  —  15  imits  from  O. 

points  on  the  semicircles,  is  level,  and  lies  at  a  constant  height 
sin  ft  decimeters  above  the  drawing  board.  That  is,  the  tie 
wire  that  connects  all  points  of  circular  angle  ft,  measured  from 
O  X  positively  towards  OY ,  lies  at  the  uniform  height  sin  ft 
decimeters  above  the  drawing  board. 

A  plan  view  of  the  model,  taken  from  a  point  on  the  0  Z 
axis,  +  15  units  above  0,  is  given  in  Fig.  47.  It  will  be  seen 
that  each  semicircle  forms  an  ellipse,  when  projected  on  the 
Isase  plane  X  O  Y.  The  semi-major  axis  of  this  ellipse  has 
length  cosh  ft,  where  ft  is  the  hyperbolic  angle  corresponding 
to  that  semicircle.     The  semi-minor  axis  is, — 

cosh  ft  s'm  Oa  =  cosh  ft  tank  ft  =  sink  ft 
from  the  well  known  relation  that  exists  between  a  hyperbolic 
angle  and  its  gudermannian  circular  angle;  namely, — 

sin  ftj  =  tanh  ft 
All  of  these  ellipses  have  the  same  center  of  reference  0 
Any  such  system,  having  semi-major  axes  cosh  ft,  and  semi- 
minor  axes  sinh  ft,  are  well  known  to  be  confocal,  and  the  foci 
must  lie  at  the  points  +1  and  — i  in  the  X  O  K  plane,  or  the 
points  in  which  the  innermost  circle  cuts  that  plane. 

PROCEDURE   FOR    PROJECTING   COSH    (  ±  ft  ±    /ft) 

Thus  premised,  the  process  of  finding  the  cosine  of  a  com- 
plex hyperbolic  angle  ft  +  /ft;  that  is,  the  process  of  finding 
cosh  (ft  +  /ft)  is  as  follows : 

Find  the  arc  C  E,  Fig.  45,  from  C  z=  -\-i  along  the  rect- 
angular hyperbola  C  E  H,  which  subtends  ft  radians.  The 
hyperbolic  sector  comprised  between  the  radius,  O  V,  the  hyper- 


bolic arc,  and  the  radius  vector  0  E,  on  this  arc  from  the 
origin  O,  will  then  include-^  sq.  dm.  of  area.  Drop  a  vertical 
perpendicular  from  E  onto  O  X.  It  will  mark  off  a  horizontal 
distance  O  D  equal  to  cosh  ft.  Proceed  along  the  circle  which 
rises  at  D,  in  a  positive  or  counterclockwise  direction,  through 
ft  circular  radians,  thus  reaching  on  that  circle  a  point  G  wheise 
elevation  above  the  drawing  board  is  sin  ft  decimeters.  The 
area  cnclo'^ed  by  a  radius  vector  from  the  oriein  O  on  the  circle, 
followed  between  the  axis  O  C  and  the  circular  curve,  will  be 

—  cosh  'ft  sq.  dms. 

From  G,  drop  a  vertical  plummet,  as  in  Fig.  46,  on  to  the 
drawing  board.  In  other  words,  project  G  orthogonally  on  the 
plane  X  O  Y.  Let  g  be  the  point  on  the  drawing  board  at 
which  the  plummet  from  G  touches  the  surface.  Then  it  is 
easily  seen  that  Og  on  the  drawing  board  is  the  required  mag- 
nitude and  direction  of  cosh  (ft  +  /ft),  in  decimeters,  with 
reference  to  O  X  as  the  initial  line  in  the  plane  X  O  Y.  It  may 
be  read  off  either  in  rectangular  coordinates  along  axes  O  X 
and  O  y  on  a  tracing  cloth  surface  as  shown  in  Fig.  47,  or  'n 
polar  coordinates  printed  on  a  sheet  seen  through  the  tracing 
cloth. 

If  the  circular  angle  ft,  i.  e.,  the  imaginary  hyperbolic  angle 
/ft,  lies  between  ir  and  2t  radians,  (in  quadrants  3  and  4), 
the  point  G  will  lie  on  the  under  side  of  the  plane  X  O  Y,  and 
the  projection  onto  g  in  that  plane  must  be  made  upwards, 
instead  of  downwards. 

If  the  hyperbolic  angle  whose  cosine  is  required  has  a  nega- 
tive imaginary  component,  according  to  the  expression  cosh 
(ft  —  /ft),  then  starting  from  the  projected  point  Z),  we  must 
trace  out  the  circular  angle  in  the  negative  or  clockwise  direc- 
tion, as  viewed  from  the  front  of  the  model. 


|l||l|||||||l|l||iy|||||iili|||||li 

1  mtti  BMi 



;;;;i-:|.:'j;!;|4:|;;i:i;i:;;i|^' 

Y 

■■'i"i-ii'i''8i''  J'S'STi' 

m 

=£=»    ■ 

[444-;2!o|44-ii5|44  iid-j  -o'54i4o444*-ci.q4>ji  d-frnisl-Lp-^  (H-(+h 

""TXt      'iJ-     It     X      A-      ---    --U-      --iJ4--- 

'   1  .      II 

AH-    -  -  +  -1-   -Hi-t+  -t~     0                  c          iD-t-  . . 

mil" 

FIG.  47 — PLAN   VIEW  OF   MODEL 

From  a  point  on  the  O  2  axis,  15  units  from  O. 

If  the  real  part  of  the  hyperbolic  angle  is  negative,  accord- 
ing to  the  expression  cosh  (—ft  ±  /ft)  ;  then  since  cosh  — {Bt 
=!=  /ft)  =  cosh  (ft  =F  /ft),  we  proceed  as  in  the  case  of  a  posi- 
tive real  component,  but  with  a  change  in  the  sign  of  the 
imaginary  component. 

The  operation  of  tracing  cosh  (  ±  *t  ±  /ft)  on  the  X  Y 
plane,  thus  calls  for  two  successive  orthogonal  projections  onto 
that  plane;  namely  (i)  the  projection  corresponding  to  cosh 
(±  ft)  as  though  /ft  did  not  exist,  and  then  (2),  the  projection 
corresponding  to  cosh  /ft  =  cos  ft  independently  of  ft,  except 
that  the  radius  of  the  circle,  and  its  plane,  are  both  conditioned 
by  the  magnitude  of  ft. 

If  we  trace  the  locus  of  cosh  (ft  ±  /ft),  where  ft  is  held 
constant,  it  is  evident  from  Fig.  47  that  we  shall  remain  on  one 
circle,  which  projects  into  the  same  corresponding  ellipse  on 
the  X  Y  plane.  That  is,  the  locus  of  cosh  (ft  i  /ft)  with  ft 
held  constant,  is  an  ellipse,  whose  semi  major  and  minor 
diameters  are  cosh  ft  and  sinh  ft  respectively.  If,  on  the  other 
hand,  we  trace  cosh  (  ±  ft  -f-  /ft)  with  ft  held  constant,  we 
shall  run  over  a  certain  tie  wire  bridging  all  the  circles  In  the 
model,  which  tie  wire  is  sin  ft  dm.  above  the  board,  and  it"s 
projection  on  the  board.  In  the  plane  X  y  of  projection.  Is  part 
of  a  hyperbola. 

PROCEDURE  FOR  SINH    (ft  -f-  /ft) 

It  would  be  readily  possible  to  produce  a  modification  of 
this  model  here  described,  which  would  enable  the  sine  of  a 
complex  angle  to  be  projected  on  the  X  K  plane  following  con- 
structions already  referred  to.  The  transition  to  a  new  model 
for  sines  is,  however,  unnecessary.    It  suffices  to  use  the  cosine 


94 


HYPERBOLIC  FUNCTIONS 


model  here  described  in  a  slightly  different  way.    One  has  only 
to  recall  that 


sink 


e  =  —i  cosh\9  +7-7) 


or 


sink  (9.  +  M)  =  — y  cosh  [ei+j(6'i  +  -7)] 


Consequently,  in  order  to  find  the  sine  of  a  complex  hyperbolic 
angle,  we  proceed  on  the  model  as  though  we  sought  the  cosine 

TT 

of  the  same  angle,  increased  by  -7- radians  or  one  quadrant,  in 

the  imaginary  or  circular  component.  We  then  operate  with 
— }  on  the  plane  vector  so  obtained;  i.  e.,  we  rotate  it  through 
one  quadrant  in  the  X  Y  plane  and  in  the  clockwise  direction. 
An  equivalent  step  is,  however,  to  rotate  the  X  and  Y  axes  of 
reference  in  that  plane  through  one  quadrant  in  the  reverse  or 


positive  direction."*That  is,  we  may  omit  the  — j  operation.  It, 
in  dealing  with  sine  projections,  we  treat  O  y  as  an  O  X  axis, 
and  — O  X  as  an  O  F  axis,  or  read  off  the  projections  on  the 
X  Y  plane  to  the  — Y  O  Y  axis  as  initial  line. 

The  only  difference,  therefore,  between  projecting  the 
cosine  and  the  sine  of  a  complex  hyperbolic  angle  in  the  model, 
is  that  in  the  latter  case  the  circular  component  is  increased  by 
one  quadrant  and  the  projected  plane  vector  is  read  off  to  the 
O  y  reference  axis  as  initial  line.  The  model  thus  gives  the 
projection  of  either  cosh  (  rt  Sii  ]'$•)  or  sinh  (  ^0i  '-  j'S-.) 
within  the  limits  of  +1-4  ^nd  — 1.4  for  0i,  and  for  fij  between 
the  limits  +  <^  and  —  a  .  For  accurate  numerical  work,  refer- 
ence would,  of  course,  be  made  to  the  charts  and  tables  of  such 
functions  already  published,  and  which  enable  such  functions 
to  be  obtained  either  directly  or  by  interpolation,  for  all  or- 
dinary values  of  Si  and  62. 


CHAPTER  XI 
PERFORMANCE  OF  LONG  TRANSMISSION  LINES 

(RIGOROUS  SOLUTION  BY  HYPERBOLIC  FUNCTIONS) 


AS  STATED  in  the  discussion  of  the  convergent 
series  solution,  the  performance  of  an  electric 
circuit  is  completely  determined  by  its  physical 
characteristics; — resistance,  reactance,  conductance  and 
capacitance  and  the  impressed  frequency.  These  five 
quantities  are  accurately  and  fully  accounted  for  in  the 
two  complex  quantities. 

Impedance  Z  =::  R  -^  jX 
Admittance  Y  =  G  +  jB 

Having  determined  the  numerical  values  for  these 
two  complex  quantities,  no  further  consideration  need 
be  given  to  the  physical  quantities  of  the  circuit  or  to 
the  frequency. 

In  the  hyperbolic  theory  the  circuit  is  said  to  sub- 
tend a  certain  complex  angle,  6  =  VzK.  This  quantity 
represents  in  a  sense  the  electrical  length  of  the  circuit. 
The  numerical  value  of  this  angle  6  is  expressed  in 
hyperbolic  radians.  If  the  circuit  is  very  long  electric- 
ally the  numerical  value  of  the  angle  will  be  compara- 
tively large.  Conversely,  if  the  circuit  is  electrically 
short,  it  will  be  comparatively  small.  The  numeiical 
value  of  the  angle  6  is,  therefore,  a  measure  of  the  elec- 
trical length  of  the  circuit  and  an  indication  of  how 
much  distortion  in  the  distribution  of  voltage  and  cur- 
rent is  to  be  expected  as  an  effect  of  the  capacitance  and 
leakance  of  the  circuit. 

In  order  to  give  an  idea  of  the  extent  of  the  varia- 
tion in  the  complex  6  and  its  functions  cosh  6  and  sinh  6 
for  power  transmission  circuits  o£  various  lengths  cor- 
responding to  25  and  60  cycle  frequencies  approximate 
values  have  been  calculated,  as  shown  in  Table  O. 

This  tabulation  indicates  that  for  circuits  of  f -om 
100  to  500  miles  in  length,  operated  at  frequencies  of  25 
and  60  cycles,  the  complex  hyperbolic  angle  of  the  cir- 
cuit (which  is  a  plane-vector  quantity)  has  a  maximum 
modulus,  or  size  of  0.41  for  25  cycles  and  of  1.05  for  60 
cycles.  It  has  an  argument,  or  slope,  lying  between  70 
and  78  degrees  for  25  cycles  and  between  80  and  85 
degrees  for  60  cycles. 

In  the  convergent  series  solution,  the  three  so-cr.Iled 
auxiliary  constants  A,  B  and  C  determine  the  perfonn- 
ance  of  the  circuit.  These  three  auxiliary  constants  are 
simply  expressions  for  certain  hyperbolic  functions  of 
the  complex  hyperbolic  angle  B  of  the  circuit. 

Thus 

A  T=  cosh  0 

r  sinh  6 


B  =  sinh 


>s 


7.i 


=  Z' 


C  =  Sinn  e  — r^  —  ' z — 

Z  * 


ADDITIONAL  SYMBOLS 

In  addition  to  the  symbols  previously  listed,  the 
following  will  be  employed  in  the  hyperbolic  treatment. 

a  =  Linear  hyperbolic  angle  expressed  in  hyps 
per  mile.  It  is  a  complex  quantity  consist- 
ing of  a  real  component  at  and  an  imaginary 
component  oc  i.  It  is  also  known  as  the  at- 
tenuation constant  or  the  propagation  con- 
stant of  the  circuit. 

Oi  =  The  real  component  of  the  linear  hyperbolic 
angle  a,  expressed  in  hyps.  It  is  a  measure 
of  the  shrinkage  or  loss  in  amplitude  of  the 
traveling  wave,  per  unit  length  of  line 
traversed. 
«i  =  The  imaginary  component  of  the  linear 
hyperbolic  angle  a,  expressed  in  circular 
radians.  It  is  a  measure  of  the  loss  in  phase 
angle  of  the  traveling  wave,  per  unit  length 
of  line  traversed. 

•  =  The  complex  hyperbolic  angle  subtended  by 
the  entire  circuit,  expressed  in  hyps.  It 
differs  from  a  in  that  it  embraces  the  entire 
circuit,  whereas  a  embraces  unit  length  of 
circuit  (in  this  case  one  mile),  $  =:  a  XL, 
where  L  is  the  length  of  the  circuit  expressed 
in  miles. 

ft  =  The  real  component  of  the  complex  hyper- 
bolic angle  of  the  circuit  expressed  in  hyps, 
and  defines  the  shrinlcage  or  loss  in  ampli- 
tude or  size  of  a  traveling  wave,  in  travers- 
ing the  whole  length  of  the  line. 

e,  =:  The  imaginary  component  of  the  complex 
hyperbolic  angle  of  the  circuit  expressed  ia 
circular  radians,  expressing  the  loss  in  phase 
angle  or  slope  of  the  traveling  wave,  in 
traversing  the  whole  length  of  line, 
e  =  2.7182818  which  is  the  base  of  the  Napierian 
system  of  logarithms.     Logw    =  0.4342945. 

6,  =  Position  angle  at  sending  end. 

6,  =  Position  angle  at  receiving  end. 

0p  =  Position  angle  at  point  P  on  a  circuit 
S  z=  Impedance  load  to  ground  or  zero  potential 
at  receiving  end  line,  in  ohms  at  an  angle. 

».  =y'-^=  Surge  impedance  of  a  conductor  in  ohms  at 


y 

y.  =  —  =  Surge  admittance  of  a  conductor  in  mhos  at 


an  angle. 
Surge  adi 
an  angle. 


TABLE  0~GENERAL  EFFECT  OF  DISTANCE  AND 
FREQUENCY  UPON  THE  COMPLEX  HYPER- 
BOLIC ANGLE  AND  ITS  FUNCTIONS 


LENQTM  Of 
OlROUrT 
(MILES) 

z 

r 

«-/Iv 

COSHt 

SMH» 

25       CYCLES 

/oo 

-f,^3,  .  fP 

0.000  230  Lfi?' 

o.cc«»4\ii«; 

n.  inijir 

0.991012' 

<jL<fl/7a» 

zoo 

f  O.A , ,  st 

0.000  ^30\SS' 

O-OJ^tiVilfi" 

0.1*  lit 

o.9tLa±.r 

0,190X1 

300 

f09>  1  S 

0.000  67012c 

0.07303\^^£ 

o.nliS 

o.v*^,p.y 

n.iylffl 

40O 

'-♦  3  ,  ,  jfc" 

0.000  900  i2Cf 

0.  /2  »70  VjEA." 

0.3*  i2i? 

0.9^  if.s* 

0.3523? 

■500 

'SAL  2.* 

0.00/  /oo  ]S^ 

0.  '  7/60  \/J7' 

o.-ti  at 

o.9,li.i^ 

0.3* /r9; 

6   0       C 

Y    C    L    E 

s 

3.00 

■ftsLiL' 

0.00/  osow^ 

0./73»ia  2' 

O.tiLL 

l:"i£i- 

'  ^'^?yf^' 

30O 

Z-^4179* 

a. 001  iootSS! 

0.3?O.»0Ui  i' 

0.13  11: 

O.tf     A.  J* 

O.i0  LMSl 

400 

^^t.lliL'. 

0.002  ISOISS.' 

0.70090\^JS^ 

O.ULt 

0.  A  7     X.f 

0,7*  IS^ 

■500 

'i-OTULO- 

0.002  700122" 

).osU£ 

0.5/     lii." 

0.97  LK27 

These  values  are  but  roughly  approximate  to  illustrate  the 
general  effect  for  certain  circuits. 


96 


HYPERBOLIC  SOLUTION   FOR  LONG   LINES 


DETERMINATION   OF  THE  AUXILIARY   CONSTANTS 

It  was  shown  in  Chart  XI  how  values  for  the 
auxihary  constants  A,  B  and  C  may  be  determined 
mathematically  by  convergent  series  form  of  solution, 
u?ing  problem  X  as  an  example.  Chart  XVI  gives  in- 
formation as  to  how  these  same  auxiliary  constants  may 
be  determined  by  the  use  of  real  hyperbolic  functions. 

The  solution  for  the  auxiliary  constants  by  real  hy- 
p-crbolic  functions  is  given  completely  for  problem  X 
in  Chart  XVI.  Vector  diagrams  are  given  to  assist  in 
following  the  solution.  In  the  solution  for  the  auxiliary 
constants  by  convergent  series,  the  operations  were 
carried  out  by  aid  of  rectangular  co-ordinates  of  the 
complex,  or  vector  quantities.  In  Chart  XVI,  the  op- 
erations are  to  a  large  extent  carried  out  by  the  aid  of 
polar  co-ordinates.  In  the  case  of  convergent  series, 
most  of  the  operations  consist  of  adding  the  various 
terms  of  the  series  together.     As  addition  and  subtrac- 


tion of  complex  quantities  can  be  most  readily  carried 
out  when  expressed  in  rectangular  co-ordinates,  this 
form  of  expression  is  used  for  the  convergent-series 
solution.  On  the  other  hand,  powers  and  roots  of  com- 
plex quantities  are  most  readily  obtained  by  polar  co- 
ordinate expression.  In  the  solution  by  real  hyperbolic 
functions  Chart  XVI,  operations  for  powers  and  roots 
predominate,  and  for  this  reason  polar  expressions  have 
been  quite  generally  employed.  The  solution  by  real 
hyperbolic  functions  is  briefly  this : — 

The  impedance  Z  and  the  admittance  Y  are  first 
set  down  in  complex  form  and  their  product  obtained. 

square  root  of  this  product  gives  the  complex  angle 
6  ==V^zyof  the  circuit.  This  angle  is  then  exprested 
in  rectangular  co-ordintes  as  6^  -)-  /  6^  for  the  purpose 
of  determining  the  numerical  value  of  its  real  part  6^ 
(expressed  in  hyps)  and  its  imaginary  or  circular  part 
^2  expressable  in  circular  radian,^.     This  circulnr  par'  0^ 


CHART  XVI— RIGOROUS  SOLUTION    FOR  AUXILIARY  CONSTANTS  OF  PROBLEM   X 

BY  REAL  HYPERBOLIC  FUNCTIONS 

CHARACTERISTICS   OF  CIRCUIT 

LENGTH  300  MILES.  CYCLES  60. 

CONDUCTORS  — 3  »  ODD  STRANDED  COPPER. 
SPACING  OF  CONDUCTORS    10X10X20  FEET. 
EQUIVALENT  DELTA  SPACING  =  l2.e  FT. 

LINEAR  CONSTANTS  OF  CIRCUIT 

TOTAL  PER  CONDUCTOR 
R  -  0.360  X  300  =  106  OHMS  TOTAL  RESISTANCE  AT  26°  C. 
X=  0.830X300  =  249  OHMS  TOTAL  REACTANCE. 
Q  =  6.21  X  300  X  10"^  =  .001563  MHO  TOTAL  SUSCEPTANOE. 
G  =     0  X  300  =  0  MHO  TOTAL  CONDUCTANCE. 
g  =  (IN  THIS  CASE  TAKEN  AS  ZERO). 


-ERN><ai * 

VOLTAGE  AT  RECEIVING-END  — 


VOLTAGE   DIAGRAM 


SOLUTION    FOR  0  = 

=  N2Y 

ez  =  TAN-'^  =  TAN-l2^  =  67*  8'  8' 
2  =  R  +  jX  =  IOS  +  j249 

7     i^ 

4i 

=  11062  +  24g2  ^  270.233 
-    270.233  /67"  8'  8" 

eY-TAN-lB-TAN-l°-°°^S"        90- 
Y  =  G  +  jB  =  0  +  ;  0.001663 

=  V  0^  +  0.0015632=  0.001563 
=  0.001563  1  90" 

Y 

®ZY  =  9z  +  ®Y  =  67°  8'  8'  +  90°  =  1  57°  8'  8' 
ZY  =  270.233  X  0.001  563  =  0.4223746 
=  0.4223746  Vl6''°  8' 8' 

\ 

e^. 

e=VZY'=  0.6499035  As"  34' 4'  HYP. 
=  O.I288l7+j  0.637009  HYP. 

-     "-(a.+jag) 

// 

i^^" 

- 

/|/Z      /i;270.233  /e7-8'8' 
IfY       1/    0.001663  |_90° 

=  7l72893\22"6l'  6^" 

=      4l5.806\ll°26' 68" 

WAVE  LENGTH 

92=   +j  0.637009  HYP. 
0^=0^^9  =  0.002,2333 

WAVE  LENGTH  =^^=§^||ii||5I2  =  2969  MILES 

SOLUTION    FOR  (A) 

(A)=    COSH     YZY=(C0SH  e,  COS  Oj  +  j  SINHO,  SlNOj) 


61  =  0.128817  HYP        0, 


.360° 


X  0.637009  =  36°  29'  52' 


LOG  COSH  e|=  0.003594 

LOG  COS  B2=T. 905194 

LOG  a,  =T. 908788 

a,=  0.81058 


LOGSINH  0|  =  I  .111172 

LOG  SIN  eo=T.774359 

LOG  32=7.885531 


r 


=  0.07683 


0.81066 +j  0.07683 
(A) -0.8142   /5°  24'  62' 


SOLUTION    FOR(B) 

(B)  •=  yf  SINH  fZY  =Y|^(SINH  e,  COS  92+ j  COSH  0,  SIN  a^) 
LOG  SINH  e,  =7.111172  LOG  COSH  0,  =0.003694 


LOG  COS   82=j  .906194 
I  .016366 


LOG  SIN  On  =1.774359 
7.777963 
SINH  0|  COS  02=0.10383       COSH  0,  SIN  02="O.68873 

TAN-I  °-69973  =  /  oq.  .  „,  .„, 
0. 1  0383  ^-52_L2_*2_ 

SINH  e,  COS  02+ j  COSH  0|   SIN  02  = 

=  O.I0383+;0.69973 

=  0.60866   /80°  10' 40' 


ff= 


'|-  =  4I5.B05   \ir  25'56' 

=  415.805    \l  l°25'  56*  X  0.60866   /80°I0'40' 
(B)  =  253.08     /68°44'44' 

SOLUTION    FOR  (C) 


(C)=    73  SINH  l/ZY 


—  ,,,„„      ^     ,                 X  0.60865   /80°  10' 40' 
416.805   \ll°  26'  58'  ' 

-  0.002405  /l  1°  26'  56'  X  0.60865    /80°  1 0'  40' 
(C)=  0.00 1 464 \   91°  36'  36' 


As  a  check  against  possible  serious  errors  in  the  calculations,  the  calculated  values  may  be  compared  with  values  read  front 
the  Wilkinson  Charts.     The  above  results  check  exactly  with  those  obtained  by  convergent  series.     (See  Chart  XI). 


HYPERBOLIC  SOLUTION   FOR   LONG   LINES 


97 


is  converted  to  degrees  by  multiplying  by  57°  .29578. 
The  hyperbolic  cosine  and  sine  of  this  complex  angle  are 
next  obtained  by  the  aid  of  logarithms  of  the  functions 
of  the  component  parts  of  the  hyperbolic  complex  angle 
0.  The  equation  for  cosh  0  and  sinh  6  is  given  just 
above  the  solution.  With  a  vievtr  of  eliminating  the 
necessity  of  calculation  for  each  complex  angle,  cosh  6 
and  sinh  0,  Dr.  Kennelly  has  prepared  tables  and  charts 
from  which  these  two  functions  (and  others)  may  be 
obtained  directly,  thus  very  materially  shortening  the 
solution  by  hyperbolic  functions.  Since  complex  angles 
have  two  variable  components  (O^  +  ;  6^)  tables  of 
functions  of  such  angles  would  have  to  be  quite  exten- 
sive in  order  that  the  steps  for  which  values  for  the 
functions  are  given  be  not  excessive.  Although  tables 
of  functions  of  complex  angles  are  not  as  complete  as 
is  desired  they  are  a  great  help  in  the  solution  of  ordi- 
nary power  circuits.  Functions  corresponding  to  angles 
lying  between  the  values  for  angles  in  these  tables  may 
readily  be  approximated  by  simple  proportion,  giving 
values  sufficiently  accurate  for  ordinary  power  trans- 
mission circuits.  They  have  been  calculated  in  Chart 
XVI  for  the  purpose  of  illustrating  such  procedure  and 
also  as  a  high  degree  of  accuracy  was  here  desired  for 
the  purpose  of  illustrating  the  agreement  of  the  results 
as  obtained  by  different  rigorous  methods.  Ordinarily 
these  values  would  be  taken  from  tables. 

SOLUTION  BY  NOMINAL  tt  METHOD 

By  this  method,  in  place  of  considering  the  admit- 
tance of  the  circuit  as  being  distributed  (as  it  is  in  the 
actual  circuit)  it  is  based  upon  the  assumption  that  the 
total  conductor  admittance  may  be  lumped  at  two  points, 
one  half  being  placed  at  each  end  of  the  circuit.  Such 
an  artificial  circuit  is  known  as  a  "nominal  -ir"  circuit 
since  the  nominal  values  of  impedance  and  admittance 
are  ascribed  to  this  circuit.  On  the  above  assumption, 
the  current  per  conductor  is  the  vector  sum  of  the 
receiving  end  load  and  the  receiving  end  condenser  cur- 
rents. The  sending  end  current  is  the  vector  sum  of 
the  conductor  and  the  sending  end  condenser  currents. 
The  performance  of  such  a  circuit  may  be  determined 
either  graphically  or  mathematically. 

If  the  circuit  is  not  of  great  electrical  length,  (say 
not  over  100  miles  at  60  cycles  or  200  miles  at  25 
cycles)  the  performance  of  the  corresponding  nominal 
w  circuit  will  not  be  materially  different  from  that  of 
the  actual  circuit  having  distributed  constants  which  it 
imitates.  If,  however,  the  circuit  is  of  great  electrical 
length  the  performance  of  the  nominal  ir  circuit  no 
longer  closely  imitates  the  performance  of  the  actual 
circuit  which  it  represents,  owing  to  an  error  due  to  the 
lumpiness  of  the  artificial  circuit.  Dr.  Kennelly  has 
shown  that  by  making  certain  modifications  in  the  linear 
or  fundamental  constants  for  the  impedance  and  admit- 
tance of  the  nominal  v  circuit,  the  lumpiness  error  will 
vanish,  so  that  the  artificial  circuit  will  then  truly  re- 
present at  the  terminals  the  behavior  under  steady  state 


operation,  taking  distributed  admittance  into  account. 
Such  a  corrected  artificial  circuit  is  known  as  the 
"equivalent"  w  circuit,  because  it  then  becomes  externally 
equivalent  to  the  actual  circuit,  having  distributed  con- 
stants, in  every  respect. 

The  complex  numbers  which  must  be  applied  to 
the   impedance,   Z  and   the   admittances, —   and  

of  the  nominal  ir  circuit  in  order  to  correct  these  nom- 
inal values  into  the  equivalent  circuit  are  called  the  cor- 
recting factors  of  the  nominal  ir  circuit.  The  nominal 
values   of   the   impedance   Z   and   the   admittances  — 

of  the  circuit  must  be  multiplied  by  these  vector  correct- 
ing factors  in  order  to  convert  them  into  the  "equivalent" 

values;  thus: — 

s\nh  e 


2     ~    2 


e 

tank  e/ 2 
e/2 


Where  0  =  yJZY  is  the  hyperbolic  complex  angle  sub- 
tended by  the  circuit. 

Complete  tables  of  hyperbolic  functions  are  not  al- 
ways available;  then  again,  many  engineers  have  a  nat- 
ural aversion  to  the  use  of  such  functions.  In  order 
to  avoid  these  objections  as  well  as  to  simplify  calcula- 
tions. Dr.  Kennelly  has  charted  these  "correcting 
factors"  for  hyperbolic  complex  angles  up  to  0 
=  i.o  radian  in  steps  of  o.oi  in  size  and  i  degree  in 
slope.  The  writer  is  particularly  indebted  to  Dr. 
Kennelly  for  these  charts,  which  are  reproduced  here- 
with for  the  first  time,  as  Charts  XVIII,  XIX,  XX  and 
XXI.  It  is  believed  that  the  use  of  these  charts  will 
greatly  simplify  the  calculation  of  the  performance  of 
electric  power  transmission  circuits  by  hyperbolic  func- 
tions. They  enable  the  vector  values  of  these  ratios  to 
be  read  to  at  least  three  decimal  places  in  sizes  and  to 
two  decimal  places  in  slope,  and  their  availability  makes 
the  use  of  tables  of  hyperbolic  functions  unnecessary. 
The  corrected  conductor  impedance  Z'  is  the  same  as 
the  familiar  auxiliary  constant  B. 

EQUIVALENT  ir  SOLUTION  FOR  PROBLEM  X 

The  solution  for  problem  X  by  the  equivalent  r 
method  is  given  in  Chart  XVII.  At  the  top  of  the  sheet 
are  two  diagrams,  one  a  diagram  for  one  conductor  of 
the  circuit  of  problem  X  and  the  other  a  corresponding 
vector  diagram  of  the  currents  and  the  voltages  at  both 
ends.  The  numerical  values  of  the  angles  and  the  quanti- 
ties pertaining  to  problem  X  are  placed  upon  the  two 
diagrams  for  the  purpose  of  assisting  in  following  the 
mathematical  solution. 

The  physical  properties  of  the  circuit  are  first  set 
down,  its  linear  constants  obtained  from  the  tables  of 
constants  and  multiplied  by  the  length  of  the  circuit  to 
obtain  the  total  values  per  conductor.  The  next  pro- 
cedure is  to  calculate  the  hyperbolic  angle  0  of  the 
circuit.  To  do  this  the  impedance  and  the  admittance 
of  the  circuit  are  set  down  as  complex  quantities  in  the 
form  of  polar  co-ordinates  and  multiplied  together  by 
multiplying  their  slopes  and  adding  their  angles.  The 
square  root  of  the  resulting  vector  is  obtained  by  tak- 


HYPERBOLIC  SOLUTION  FOR  LONG  LINES 


CHART  XVIII 
KENNELLY  CHART  FOR  IMPEDANCE  CORRECTING  FACTOR 

(FOR  ANGLES  HAVING  POLAR  VALUES  BETWEEN  0  AND  0.40) 


READ  POLAR  VALUE  "SIZE'OF  CORRECTING  FACTOR  HERE 


COPYRIGHT  1920  A.  E.  KENNELLY 


To  find  the  vector  "correcting  factor"  corresponding  to  any  complex  line  angle  0,  of  a 
circuit,  the  angle  8  is  expressed  in  polar  form  with  the  slope  in  fractional  degrees.  The 
correctmg  factor  as  read  from  the  chart  will  be  in  polar  form  with  its  slope  in  fractional 
degrees.     Consult  Table  P  for  rapid  conversion  to  minutes  and  seconds.     For  example; — 

e  —  0.3  768°.  correcting  factor  =  a9893  /o°  .Co  =  0.9893  /o°36'oo" 

0  =  a2l5  /8o°  .5,  correcting  factor  =  o.fKjs;  /o°  .149  —  a9927  /o°o8'56" 


HYPERBOLIC  SOLUTION  I'OR  LONG  LINES 

CHART  XIX 
KENNELLY  CHART  FOR  IMPEDANCE  CORRECTING  FACTOR 


V9 


(FOR  ANGLES  HAVING  POLAR  VALUES  BETWEEN  0  40  AND  1 .0) 


00 


CO 


READ  POLAR  VALUE      SIZE"  OF  CORRECTING  FACTOR  HERE 
O 

a> 
6 


oo 


00 
00 


00 


—   w 


■CO 


a> 


03 


o> 


a>   o 
o>   o 

d   — 


COPYRIGHT   1920  A  E  KENNELLY 


To  find  the  vector  "correcting  factor"  corresponding  to  any  complex 
line  angle  0,  of  a  circuit,  the  angle  6  is  expressed  in  polar  form  with  the 
slope  in  fractional  degrees.  The  correcting  factor  as  read  from  the  chart 
will  be  in  polar  form  with  its  slope  in  fractional  degrees.  Consult  Table 
P   for  rapid  conversion  to  minutes  and  seconds.     For  example:— 

0  =  0.8  /62°.  correcting  factor  =  0.943  /.s"  .TO  =  0.943  /■;°ii'24'' 
e  =  0.6499  /78"  .57,  correcting   factor  =  0.9365   /i°  .6t   =  a9365 
/i°.^6'.^6" 


leo 


HYPERBOLIC  SOLUTION   FOR  LONG   LINES 


CHART  XX  KENNELLY CHART 
FOR  ADMITTANCE  CORRECTING   FACTOR 

(FOR  ANGLES  HAVING  POLAR  VALUES  BETWEEN  0  AND  0.20) 


90°- 


0.05 
0.06 


READ     POLAR  VALUE      SIZE' 


OF    CORRECTING  FACTOR  HERE 

COPYRIGHT  1920  A.  E.  KENNEULY 


HYPERBOLIC  SOLUTION  FOR  LONG  LINES 


lOI 


CHART  XXI   KENNELLY  CHART 
FOR  ADMITTANCE  CORRECTING  FACTOR 

(FOR  ANGLES  HAVING  POLAR  VALUES  BETWEE;N  0.20  AND  0.50) 

r80°T0° 


-2.r 


u 

c 

111 

I 

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0 

t- 

0 

< 

k. 

0 

Z 

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u 

u 

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K  u  >> 

V 

oc 

VJiXl 

J3 

0 
0 

fi-S-S 

v^ 

ll 

o-c'o 

0 

S  S-? 

A 

111 
n 

^^'^ 

^ 

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0) 

to  a 
r  fo 
then 

(A 

u 

ding 
pol 
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« 

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K 

c.E  u 

u 

0        c 

s-s- 

a 

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ress 
the 

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a 

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V 

0.2- 

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4: «  c 

^-" 

c   ho  u 

recti 

e  an 

Th 

0 

cor 
,  th 
ees. 

b 

c« 

the  voctor 

of  a  circui 

ctional  dcg 

u 

CI 

C 

•0*2 

t« 

rt 

^i;"** 

/•^•S^ 

id  = 

0  C--  «■  1  ^ 

H  «  4> 

u  0. 

— 

c  0 

V 

a  3.1 

O         19 

•o  j;  ^ 

•o 

•-  a  c 
0.9  o 

y  <J  »< 
«  ca  « 

to      c 

c  c  « 


O  t)  «i 

o  «  n 
c  « 


be  € 


u^    I/l 


READ  POLAR  VALUE  "siZE"   OF  CORRECTING  FACTOR  HERE 

COPYRIOHT  1 920  A.  E.  KENNELLY 


102 


HYPERBOLIC  SOLUTION  FOR  LONG  LINES 


ing  the  square  root  of  the  slope  and  halving  the  angle. 
The  result  is  the  hyperbolic  angle  6  of  the  circuit  ex- 
pressed in  hyps. 

The  ratio  charts  XIX  and  XXI  are  next  consulted 

,  ^,                ^.           ,        sink  0         .    tank  6/2 
and  the  correcting  values  — ^ and  corre- 

sponding to  ^e  'hyperbolic  angle  of  the  circuit  read 
off.  Having  thus  obtained  the  correcting  factors  corre- 
sponding to  this  circuit,  the  linear  impedance  Z  and 
linear  admittance  Y  per  conductor  are  multiplied  re- 
spectively by  the  sinh  and  the  tanh  correcting  factors. 

If  the  circuit  under  consideration  is  electrically 
short  the  effect  of  these  correcting  factors  upon  the 
linear  constants  will  be  small  and  possibly  negligible 
but,  as  the  circuit  becomes  longer,  their  effect  be- 
comes increasingly  greater.  The  effect  of  the  cor- 
recting factors  for  problem  X  is  to  change  the 
linear  impedance  Z  from  270.233  /67°  08'  08"  to 
Z'  =  253.083  /68°  44'  41"  and  to  change  the 
linear  admittance  Y  from  0.001563  /90°  to  Y'  = 
0.001615512  /89°  10'  45".  In  other  words  this  circuit 
will  behave  in  the  steady  state  at  60  cycles  as  though 
its  conductor  resistance  were  reduced  from  105  to 
91.7486  ohms  and  its  inductive  reactance  reduced  from 
249  to  235.866  ohms.  Similarly  it  will  behave  as 
though  a  non-inductive  leak  of  11.571  micromhos,  has 
been  applied  to  each  condenser  in  shunt. 

In  order  to  illustrate  the  exact  agreement  in  the  re- 
sults as  obtained  by  the  equivalent  tt  method  wath  those 
obtained  by  either  the  convergent  series  or  pure  hyper- 
bolic solution,  the  ratio  values  used  for  this  problem 
were  calculated  and  not  obtained  graphically.  The  ac- 
curacy in  the  performance  resulting  from  the  use  of 
ratio  values  taken  from  the  charts  is  well  within  the 
requirements  of  practical  power  circuits.  The  mathe- 
matical solution  for  these  factors  is  given  in  Fig.  48. 

Having  determined  the  corrected  values  for  the 
impedance  Z'  and  the  admittance  Y'  which  will  produce 
exact  results,  the  remainder  of  the  solution  may  be 
carried  out  graphically  as  indicated  by  the  vector  dia- 
gram in  the  upper  right  hand  part  of  Chart  XVII  or 
mathematically  as  indicated  under  this  vector  diagram. 

EQUIVALENT  T  SOLUTION 

Dr.  Kennelly  has  shown  that  the  correcting  factors 
which  convert  the  nominal  w  into  the  equivalent  tt  of 
the  conjugate  smooth  line,  are  the  same  as  those  which 
convert  the  nominal  T  into  the  equivalent  T,  but  in  in- 
verse order; — ^that  is  the  correcting  factors  for  the 
nominal  T  line  are 

tank  61 2 


Z'  =  z- 


Y'  =  Y 


eh 

sinh  0 


Either  the  equivalent  tt  or  the  equivalent  T  solution 
may  be  used  by  applying  the  two  correcting  factors 
properly.  Usually  less  arithematical  work  will  be  re- 
quired for  the  equivalent  tt  solution. 

ELECTRICAL  CONDITIONS  AT  INTERMEDIATE  POINTS 

In  the  foregoing,  the  behavior  of  circuits  at  their 
terminals  has  been  considered.     In  some  cases   it  may 


be  desirable  to  predetermine  the  voltage  and  the  current 
at  points  along  the  circuit  between  the  terminals.  This 
may  be  particularly  desirable  in  case  of  circuits  of  great 
electrical  length  and  consequently  having  a  pronounced 
bend  or  hump  in  the  voltage  graphs  representing  the 
voltage  at  points  along  the 'ciTx;uit.  In  Fig.  21  voltage 
and  current  graphs  were  shown  for  the  circuit  of  prob- 
lem X  corresponding  to  zero  load ;  also  load  conditions. 
Accompanying  this  stated  was  the  step-by-step  method 
by  which  the  current  and  voltage  at  these  intermediate 
points  had  been  determined.  In  a  corresponding  man- 
ner the  intermediate  electrical  conditions  may  be  de- 
termined by  the  employment  of  hyperbolic  functions. 
It  is  usual,  however,  when  employing  hyperbolic  func- 
tions for  determining  the  voltage  or  the  current  at  points 
along  a  smooth  circuit,  in  the  steady  state,  to  take  ad- 
vantage of  the  following  facts  relative  to  the  variation 
in  current  and  potential  from  point  to  point  in  such  a 
circuit. 

The  potentials  of  any  and  all  points  of  such  a  cir- 
cuit are  as  the  sines  and  the  currents  as  the  cosines  of 
the  corresponding  position  angles.  This  means  that  if 
the  position  angles  corresponding  to  two  points  of  a 
smooth  circuit  in  the  steady  state  are  known,  and  the 
voltage  or  the  current  at  one  of  these  points  is  also 
known,  then  the  voltage  or  current  at  any  other  point 
will  be  directly  proportional  to  the  sine  or  the  cosine 
respectively  of  the  corresponding  position  angles.  In  a 
similar  manner,  the  impedance  follows  the  tangents,  the 
admittance  the  contagents  and  the  volt-amperes  the  sines 
of  twice  the  angles.  Herein  lies  the  beauty  of  the 
application  of  hyperbolic  functions  of  complex  angles 
for  determining  the  electrical  performance  of  electric 
circuits.  The  relationship  expressed  above  (taken  from 
Dr.  Kennelly 's  "Artificial  Electric  Lines")  are  given  in 
equation  form  below  for  ready  reference : — 
P.„         sinh  e„ 

I .   _  cosh  9p 
Z     _ 

"z7  ~ 


cosh  0.    "^""^"<:  ^ 
tanh  ».. 

coth  0p 


coth  9c 


Kv-<iv\     I  sinh  2  $p 


■  numeric  l_ 


I  Kz'-OcI     \sinh  2  Be 

Where  p  and  c  are  points  along  the  circuit,  c  being 
some  point  where  the  electrical  conditions  are  known, 
and  p  the  point  for  which  they  are  to  be  computed.  The 
vertical  lines  enclosing  the  two  parts  of  the  last  equa- 
tion are  for  the  purpose  of  indicating  that  the  "size"  of 
these  complex  quantities  are  referred  to  in  this  equation. 

POSITION    ANGLES 

Reference  has  been  made  to  the  line  as  subtending 
a  certain  complex  hyperbolic  angle  6.  Since  the  circuit 
through  the  load  also  encounters  resistance  and  react- 
ance, the  load  may  be  said  to  subtend  also  a  certain 
complex  hyperbolic  angle,  so  that  the  receiving  end  of 
the  circuit  occupies  an  angular  position  df     The  total 


HYPERBOLIC  SOLUTION  FOR  LONG  LINES 


WO 


CHART  XVII— RIGOROUS  EQUIVALENT  -t  SOLUTION  OF  PROBLEM    X 


►»*• 


CHARACTERISTICS    OF  CIRCUIT 

LENGTH.  300  MILtS.  CYCLES,  60. 

CONDUCTORS-  3  *  000  STRANDED  COPPER. 
SPACING  OF  CONDUCTORS  10  X  10  X  20  FEET. 
EQUAVALENT  DELTA  SPACING  =  l2.e  FT. 

LINEAR  CONSTANTS  OF  CIRCUIT 

FROM  TABLES  PER  MILE 
TABLE   NO.  2.  r=  0-360  OHM  AT  26' O. 
TABLE   NO.  6.X=0.830  OHM  (BY  INTERPOLATION). 
TABLE  NO.  10.  b=  6.21  X  10"^  MHO  (BY  INTERPOLATION) 
g=(IN  THIS  CASE  TAKEN  AS  ZERO). 
TOTAL  PER  CONDUCTOR 
R   -  0 .360  X  300  =  105  OHMS  TOTAL  RESISTANCE. 
X  =  0.830  X  300  =  249  OHMS  TOTAL  REACTANCE. 
B  =  6.21  X  300  X  I0"^  =  .00I663  MHO  TOTAL  SUSCEPTANOE. 
G  =     0  X  300  =  0  MHO  TOTAL  CONDUCTANCE. 


SOLUTION  FOR  HYPERBOLIC  ANGLE  8=^^17 

2=l06+j249  Y  =  0  +  j°-°OI563 

,     =  270.233  /67°e'8'  —  0.001563  I  90° 

.    ©  -=Y  270.233   /67°  8'  8' X  0.00 1  663    190° 
-yo.4223745    /I  57'  8'  8' 
=     0.6499036   /7B°  34'  4'  HYP. 

-  0.6499036    /78'.6e78  HYP. 

—  O.I288I68+J0.6370092  HYP. 

FROM   DR.  KEN  NELLY'S  CHARTS 

CHART  XIX    S!M5  _  0.9i65366  ./it 6094  -  0.9366366/1°  36' 33' 

ruiPT  yvi    TANH  0  ll  ,  , 

CHART  XXI    ^ =  1.033698    \0°.8208  =  1.033698  \0''49'I5- 

*  THESE  VALUES  WERE  CALCUUTED  IN  ORDER  TO  OBTAIN  A  HIGH 
DEGREE  OF  ACCURACY  FOR  THE  PURPOSE  OF  DEMONSTRATING  THE 
FUNDAMENTAL    ACCURACY  OF  THIS  METHOD. 

CORRECTION  OF  LINEAR  CONSTANTS 

Z'  -  270.233  / 67"  8' 8*  X  0.9366366   /l'36'33' 

-  263.083    /68''44'4I'  (WHICH  IS  AUXILIARY OONSTANTCB)) 

-  9 1 .7486  +  j  236.886  OHMS 

Y'- 0.001563    [90;  X  1 .033698    \0''49'  16' 

-  0.001616612/89°  10' 46'  MHO 

Y' 

■^  -  0.000807766   /89°  1 0'  46' 

"  0.0000  I  1 67 1  +  j  0.00080767 

-  1238    \89°J0'45'   OHMS     REACTANCE. 


CALCULATION   OF  PERFORMANCE  • 

PER  PHASE  TO  NEUTRAL 

2-6,000.  KWrn-!^50_6,4oO. 


KV-ApN-i 

-RN" 


p.  104.000  - 


t         8,000XI000„ 


R  60,046 

pPj,  -  80%  UGGINQ. 


RECEIVING-END  CONDITIONS 

Iqr— 60,046X0.000807766    /b9°  1 0'  46'  —  46.6026  / 89°  1 0'  45' 

—  0.6848  + j  48.4973  AMP. 

Ir  —  99.92  (0.90  -j  0.436)  +  0.6848+ j  48.4873 

-  80.623+ J4.8322  AMPS. 
=  80.76    /3°  C'  55'  AMPS. 

PFr- COS  3°  6' 65*- 98.85%  LEADING. 
KVVqr-  60.046  (0.6948 +j 48.4973) 

=  41.72  + J29I2.069 
KWrn=  6000  (0.B0-j0.436)+41.72  +  j29l2.07 

=  644 1. 72 +  j  296.07 

12'"  90.76  /3°  6'  65'  X  263.083  / 68°  44'  4 1 ' 
«  22970   /7l°5r36'  VOLTS 
=  716l+j2l,828,VOLTS 

SENDING-END  CONDITIONS 

EsN=e0,046  +  7l51  +  j2l828 

■=  67,197+j  21,828  VOLTS 

=  70,652  /l7°59'45'  VOLTS 
I        -  70,662X0.000807766  /89°ID'45' 


■  0.8 1 76+  j  57.0636  (TO  SUPPLY  END  VOLTAGE) 
=■  67.0696  \lori0'30'  TO  VECTOR  OF  REFERENCE 
=  -  1 6.8604  + j  64.6089 

Is-  (90.023 +j4.8322)  +  {- 16.8604+ j 64.6088) 
-  73.77  +  69.44  AMPS. 
=  94.74    /38°6I'38'  a'mPS. 

PFs=  COS  38°  61' 38'- ir  59' 46'  — 93.44*  LEADING 
K  WsN-  70.662  X  94.74  X  .9344  =.  6266  KW  PER  PHASE 
LOSS=  6255-  6400  =  865  KW  PER  PHASE 
EFF"  6400X^^3533^ 

ZERO  LOAD  CONDITIONS 


REOEIVINGrENO  VOLTAGE  =  60,046  VOLTS 
IP'  =  (0.6948 +  j  48.4973)  X  9 1.7486  -  64  +  j  4449     VOLTS 
|)('  ^  (-48.4973  +  j0.6848)X236.866— -11438+ jl64 
IR'  +  IX'  -  -11374+4613 

E<,Nn=  (60,046-ll373)+4613-48890  VOLTS      /6°24'61' 


♦The  above  results  check   with  those  obtained  by  convergent   series.     (See  Chart  XIII). 


104 


HYPERBOLIC   SOLUTION   FOR   LONG   LINES 


angle  of  the  circuit  (line  and  load)  will  be  6,  +  ^ 
=  0,.  By  similar  reasoning  all  points  lying  between 
the  receiving  and  sending  ends  of  a  line  will  occupy  or 
assume  an  angular  position  6 p.  If  that  part  of  the 
linear  angle  6  of  the  line  between  the  receiving  end  and 
the  point  p  be  designated  as  6pr,  then  the  angular  posi- 
tion of  the  point  p  will  be  ^p  =  ^r  +  dp,.  Thus,  at 
a  point  in  the  middle  of  the  line,  the  position  angle  will 
be  ^p  =  ^,  +  ^p,  =  0,  +  6/2. 
y  If  the  line  is  grounded  or  short-circuited  at  the  re- 

ceiving end,  there  will  be  no  load  containing  resistance 
and  reactance,  and  consequently  no  load  angle.  In  such 
case  6 1  =  0  and  the  distribution  of  position  angles  along 
the  line  will  be  purely  a  linear  function  of  the  total  line 
angle  6.     In  such  a  case  6^  =  0. 

Load  Conditions  —  In  Fig.  49  the  procedure  is 
shown  which  may  be  followed  for  determining  by  com- 
plex functions  of  position  angles  the  current  and  the 
voltage  vectors  at  points  25  miles  apart  along  problem 
X  circuit,  under  load  conditions. 

The  procedure  is  first  to  determine  the  complex 
angle  6,,  at  the  receiving  end  resulting  from  the  load. 
The  mathematical  determination  of  this  load  angle  is 
tedious.  Such  determination  is  given  for  problem  X 
circuit  under  stated  load  in  Fig.  49.  This  complex 
angle  $,  of  the  load  (that  is  the  position  angle  at  the 
receiving  end)  is  such  that  its  complex  tangent  equals 
the  impedance  load  S  to  ground,  or  zero  potential,  at 
the  receiving  end  of  line  (ohms  z.  )  divided  by  the 
surge  impedance  Zg  of  a  conductor  (ohms  Z-  )• 
That  is, — 

S 


tank  $T  = 


Zo 


Since  we  are  here  interested  only  in  the  ratio  be- 
tween the  load  impedance  and  the  surge  impedance,  the 
values  may  be  taken  either  per  unit  length  or  total  per 
conductor.  Although  tanh  6^  is  readily  calculated,  as 
may  be  seen  by  consulting  Fig.  49,  the  subsequent  cal- 
culation for  the  corresponding  angle  ^r  is  tedious.  After 
having  calculated  the  tanh  6,,  the  corresponding  angle  ti, 
may  be  obtained  with  sufficient  accuracy  from  a  table 
of  tangents  of  complex  angles  or,  more  readily  still, 
from  a  chart  of  such  functions.*  After  having  de- 
termined the  angle  ^r  by  consulting  a  chart  of  tangents 
of  complex  angles,  or  by  mathematical  calculation,  as 
in  Fig.  49,  the  position  angles  at  points  along  the  cir- 
cuit may  easily  and  readily  be  determined  as  follows : 

The  change  in  the  position  angle  from  point  to 
point  along  the  circuit,  due  to  the  line  impedance  and 
the  Hne  admittance  is  purely  a  linear  function  of  the 
line  angle  0.  This  is  the  case  whether  the  line  is 
grounded,  loaded  or  free  at  the  receiving  end. 

Referring  to  Fig.  49,  the  angular  position  of  the 
receiving  end,  due  to  the  load  conditions  assumed,  was 
calculated  to  be  0.48047  -f  j  1.06354.  It  is  therefore 
necessary  to  add  this  angle  to  each  of  the  linear  line 
angles  of  the  various  points  along  the  line  in  order  to 
obtain  the  position  angles   of   the   points   in   question. 


Thus  the  linear  line  angle  of  the  middle  point  of  the 
Circuit  is  0.0644084  -|-  j  0.3185046  and  adding  to  this 
the  load  angle  0.48047  +  J  1-06354  gives  0.544874  + 
J  1.3820446,  which  corresponds  with  the  entry  in  the 
tabulation  of  Fig.  51  for  the  position  angle  at  the  middle 
of  the  circuit.  In  a  similar  manner  position  angles  for 
the  load  assumed  are  readily  determined  for  points  25 
miles  apart.  Having  determined  the  position  angles  for 
the  various  points  along  the  circuit,  the  sines  and  the 
cosines  corresponding  to  these  position  angles  may  be 
approximated  closely  from  tables  or  charts  of  such  com- 
plex functions,  or  may  be  calculated  accurately  by  fol- 
lowing the  equations  at  the  lower  left  hand  corner  of 
Fig.  51.  Since  the  receiving  end  voltage  and  current 
are  known  to  be  60  046  volts  and  99.92  amperes  respec- 
tively, the  voltage  and  currents  at  all  other  points  of  this 
circuit  will  be  as  the  sines  and  cosines  of  the  corre- 
sponding position  angles.  From  the  vector  quantities 
that  have  been  assigned  to  the  voltage  and  current  at 
the  points  along  the  circuit,  the  power-factors  at  these 
points  are  readily  determined. 

The  current  and  voltage  graphs  at  the  bottom  of 
Fig.  51  were  plotted  from  values  as  determined  by  the 
use  of  functions  of  position  angles.  These  check  ex- 
actly with  similar  graphs  as  determined  by  the  Wilkin- 
son charts  and  step-by-tep  process  (See  Fig.  21). 

Zero  Load  Condition — The  procedure  which  may 
be  followed  for  determining  the  position  angles  under 
zero  load,  their  functions  and  the  corresponding  current 
and  voltage  distribution  is  the  same  as  given  above  for 
load  conditions  and  is  shown  in  Fig.  50.  In  this  case, 
however,  there  is  no  load  and  consequently  no  real 
part  to  the  load  angle.  On  the  other  hand  the  imped- 
ance of  the  load  is  infinite,  that  is  8  =  oc  so  that  0,  = 

lanh'^-y-  =•  j—r.  The  effect  of  this  supersurge  impedance 
load  at  the  receiving  end  at  zero  load  is  to  cause  a  phase 
rotation  of  90  degrees  or  one  quadrant,  /  — =  1.57080 

circular  radians.  Thus,  at  zero  load,  ^ro  =(0  +  /—  )  = 
"  +  /  1 .57080  and  this  angle  must  be  added  to  each  of 
the  linear  position  angles  of  the  points  along  the  line. 
With  the  position  angles  corresponding  to  zero  load  thus 
obtained,  and  assigned  to  the  points  along  the  circuit, 
the  voltage  will  be  found  to  follow  the  sines,  and  the 
current  the  cosines,  etc.  of  these  position  angles. 

POLAR   DIAGRAM    OF   CURRENT  VOLTAGE 

In  Fig.  52  are  shown  the  polar  graphs  of  the  volt- 
age and  the  current  for  problem  X,  corresponding  to 
load,  and  also  to  zero  load  conditions.  These  polar 
graphs  were  plotted  from  the  vector  values  for  current 
and  voltage  as  tabulated  in  Figs.  49  and  50  for  each  25 
miles  of  circuit. 

♦Such  as  that  worked  out  by  Dr.  Kennelly  and  published  by 
the  Harvard  University  Press.  The  chart  atlas  referred  to 
contains  graphs  of  complex  tangents  of  complex  angles,  and  by 
following  the  chart  in  the  reverse  from  the  usual  direction  the 
complex  angle  corresponding  to  any  complex  tangent  may  be 
read  off  directly. 


HYPERBOLIC   SOLUTION   FOR   LONG   LINES 


MS 


CURRENT  &  VOLTAGE  DISTRIBUTION 

(LOAD  CONDITIONS) 


I 

I 


0  (300  MILES)  - 


-X(MILES)- 


-  (D-X)  MILES -•• 

0pR  =  a(D-X) « 


i^ 


r 
t 

i- 


NEUTRAL 


,    ep-ep„+0R 

EpN 

I 


8ENDINQ 
END 


(D-X) 

MILES 

X 

MILES 

POSITION  ANGLE 
ep=0pR+0R 

SINH  0p 

aHE  VOLTAGE  FOLLOWS 
THIS  COMPLEX  FUNCTION) 

^PN 
VOLTS  Z_ 

COSH  0p 

(THE  CURRENT  FOLLOWS 
THIS  COMPLEX  FUNCTION) 

■p 

AMPERES  L. 

PFp 

% 

O 

300 

6^=    (.O'Si'/l" 

0.  Z4Z4  9+J0.9769  3 
=  l.00(>57  l7h°03'3S' 

60  O-fb 
/O'O'O" 

a  54294 +^0.436  3  2 
=  0.h9i,S4/38°47'lO' 

99.92 
\2S'S0'3I' 

-90.00 

25 

2.7S 

0.49/ZO +-;/.// 642 

O.ZZ4Z(,+jl.009Z 
=  1.0339    /77°2S'li' 

(>l  h70 
1  l°Z4'4l" 

0.4937Z+J0.4S937 
=  0.67364/42°59'38' 

96.64 
\2/'>38'03' 

-93.33 

SO 

ZSO 

O.SOIf\+al.lh97l 
01=    hi"  01' 10' 

0.ZO43O+ri  1.0391 
=  l.0S?O    l7S.''5Z'3h' 

{,3  /73 
/Z°49'OI' 

0.44  0  64+30.48/76 
=  0.65288/-»7''33'0»' 

93.6  6 
V7''0-#'33" 

-94.04 

7S 

ZZS 

0.5IZb7^  +  ^I.ZZZ7<f 
©2=    700  03' 3r 

O.IZZS9  +jl.0b(,3 
=  I.0SI9    /80'>li,'S9' 

H°  I3'24' 

•  0.3S692+J0.50333 
=  0.63480/52°  27'25' 

91. 0& 
\IZ'I0'Z0" 

-9S.94 

100 

ZOO 

O.SZ^^l  +3/.Z7Sa7 

%=  73' o(>' or 

O.IS9I7   ■\-jl.0909 
=  /.IOZ5  /Si'^l'SS' 

,65  770 
/5°38'20' 

0.33139  +J0.S2399 
=  0.61999  IST'4I'Z2'' 

98.94 
\6°56'/9' 

-97.60 

.  IZS 

17  S 

O.S34I4  +j/.3Z%9S 
e^     7(,°09'3%' 

=  1.1207  /83° 0/4-3' 

66  854 
/  7''04'0^ 

0.27447+JO-5436/ 
=  0.(,08/S  /63»/2'3<?' 

87.2  4 
\/<>25'02' 

-99.90 

/so 

ISO 

0.S4-4-S8  +  J/.39Z04 
Bf"     79°  II' 07" 

O.I073S+;jl.l3n 
=  /./36  8  /S^OS^-'SZ" 

(>78IS 
/  8' 31' 17' 

0.2/6/8  +0O.S(>I97 
=  0.loOZII    IhS'sYsT 

86.37 
h'19'Sl' 

-99.73 

ns 

IZS 

O.SSS6I  +^I.4-3SIZ 
02=     S2°/3'36' 

O.07908+JI.I477 
=  I.IS04  /Sh'OZ'SO" 

&8&Z6 
/9'S9'S5" 

O.IS(,i,7  +aO.S7927 
=  0.i>OOZO  /74°5/'5/' 

,    86.34 
1 10°  14' lb' 

+99.99 

xoo 

100 

O.S663S  +n'/-4S8ZI_ 
6^=    SS'/i'OS 

0.04-9  2i,+,jl.li>07 
=  l.li,l8     IST'34'II'' 

69306 
/ir30'3(,'' 

0.096  08+JO.S9S0  8 
=  0.(,0279  /80''49'42" 

86.47 
/  lb'  12' 01' 

+99- tt, 

z%s 

7S 

O.57708  +Hl.Sf-IZ9 
0^=    ■S'S'irs^ 

0.0I79S-^J  I.I707 
=  /./7OS/89°07'/3'' 

69S4Z^ 

//a-os'atr 

O.03455+JO.6O93? 
=  0.(,O9i2  /8i°4S'/S' 

87.'tS 
122'  07'3r 

+?S.7tf 

ZSQ 

SO 

O.SS78Z+3I.S943S 
0;j=    9I''ZI'04'' 

-0.014-7 1  +J/./775 
=  I.I 77S  l90''4-7:S7" 

70243 

/  I4'39'22' 

-0.02784  ■*"J0-622O7 
=  0.62270   /92°33'44" 

89.33 
/STSb'oy 

+97.32 

Z7S 

zs 

0.59SSS-^iJl.i,474h 
Bj=     94''Z3'34' 

-0.O4863+j'/./8// 

-I.IBZ/  /qs'zi'zs" 

705/7 
fli,°l7'S3" 

-0.  09073  +-J0.63  306 
=  O.63f53/98"'09'22' 

,  91.74 
/34'  ll'4l'' 

■\-9S.I7 

300 

0 

0.boqz9  +JI.700SS 
62=    97''2i-'03' 

-O.OS3^I  +J/.ISJ4 
=  1.1344   /94°03'28' 

70652 
/I7°S9'63" 

-0.IS4I6  +J  0.6422  6 
■=  0. 6  6  0  5  0 // 03*2  9'4.S' 

,    94.75 
/38'52'04» 

+  99.4i 

100 


GRAPHS  OF  CURRENT  AND  VOLTAGE  {LOAD  CONDITIONS) 


100 


90 


iso 


1  70>« 


80  a, 
ui 

80  S 

0. 
S 
70  < 
O 

60  3 


3  60 
<  60 


50 


40 


40: 


g30 
^20 


30 


10 


SENDING 
END 


26 


60 


75 


100  125  160  175  200 

DISTANCE  FROM  SENOING-END  (MILES) 


225 


260 


siNH(e|-fje2)=(SiNHe|  cos  ej+j  cosh  e,  sin  02). 
co«H(e,+je2)=(cosHe|  cosOj+jSiNHe,  sinoj). 


20  ^ 


10 


275  300 

RECEIVINa 
END 


ANGLE  AT  RECEIVING  END    0r«=O. 48047+    1.06364 
ANGLE  OF  LINE    0    =0  I2882-I-    0.63701 
0=0+0    "0.60929+    l.7006f 
ONE  QUADRANT=I.57079632  CIRCULAR  RADIANS.  S  " 

ONE  CIRCULAR  RADlAN=20e264.8062'=67*  1 7' 44.8:  ^  OC  -0.00042939+j  0.002 1 2336 


FIG.   51 — CURRENT  AND  VOLTAGE  DISTRIBUTION 

For  problem  X  by  position  angles  (load  conditions). 


io6 


HYPERBOLIC  SOLUTION  FOR  LONG  LINES 


k-- 


f— 

EsNO 


CURRENT  AND  VOLTAGE  DISTRIBUTION 

(ZERO  LOAD  CONDITION) 

0(300  MILES) 


(MILES)-  - 


■^ 


(D-X)i(MILES) >i 

0pRQ=OC  (D-X) *i 


-i 

Erno 

I 


NEUTRAL 


6po~0pRc'''0Ro 


•PNO 


SENDING 
END 


RECEIVING 
END 


(D-XJ 
MILES 

X 

MILES 

POSITION  ANGLE 

6pO~0PRO'^0RO 

SINH  0PO 

(THE  VOLTAGE  FOLLOWS 
THIS  COMPLEX  FUNCTION) 

^PNO 
VOLTS  Z_ 

COSH0PO 

(THE  CURRENT  FOLLOWS 
THIS  COMPLEX  FUNCTION) 

•  po 

AMPERES  A. 

PF-po 

% 

O 

300 

O       +J/.57080 

0j-  9o'oo'oo' 

0      +y/.ooooo 
=  1.00000  1  <7o° 

60  04L> 

0 
0    |90° 

/II'ZS'S^" 

ZS 

Z75 

0.0/073   +JI.6238S 

<9?=  fs'oz'zr 

0.OOO£7+J0.99Si,S 
=  Q.99Si>5  /srSS'o-^ 

,59  9<oS 
/o'0l'S7" 

0.  0  5  307 +J  0.0/ 070 
=  O.054\4  W°23'S8" 

,      7.82 
\9o°ol'S8'' 

0 

SO 

ZSO 

O.OZI-i-i,  +d/.i7696 

^2=  ^i'O-f'SS* 

0\0O227+j0.994i>0 
=  0.99460  /B9°S2'09'' 

£9  803 
/O"  07' £1" 

O./o&98  +  ;J0.OZI33 
=  0./08U    /ll'>Z2'48" 

/£.&Z 
\9o''03'08" 

+  00. /2 

7S 

225 

0.03S  ZO  +ai.73O04 

O.OOS/I    +J0.9878S 
~0.<7878^  /89°42'l3" 

£9  317 
/0°n'47'' 

O.ISB(oi>  +J0.O3I79 
=  0.lkl8l    ///Vr-f8" 

23.37 

\9o''o(>'oa" 

+00.32. 

lOO 

ZOO 

0.04294  +y/.7»3/3 

6>2=  loz'orsf 

0.O0  9O6+J0.97S44 
=  0.97847  /89°ZS'lsf' 

58753 
/0'3\'48" 

0.  2,IO<70+jo.04/<?? 
=  0.2)503  [jl"  IS'3S" 

,     31. OS 
\9o°  lo'zi" 

+00.6/ 

IZ5 

17  S 

O.OS3i>7+j  l.'SnbZI 
02=    /Oi'lZ'Zi" 

0.0  1409  +j  0.'J6638 
-0.<fi64'S  /89°09'J0' 

.58033 
/o°  so'  10" 

o.zi,Zi,f+jO.0£lS4Z 
-O.Zi,77(o  /  WO')'  so" 

3  8.6  6 
\90Vi,'0i," 

+^0-9? 

ISO 

ISO 

0.064^/    +J/.8S'730 
02=   IO'B°l4'£i," 

O.OZOI  S  +J0.9S/6S 
=  0.  9SIS8/8d''47'07" 

£7/£  h 
/ 1°  12'  £3" 

0.3  I  3  SO    +J0.0&  /ZO 
=0.31^70   //I'O-X'/O" 

4i>.l7 
\9o°23'4i>'' 

+  /.4Z 

175 

IZS 

0.Q7SI4    +c//.9^238 

02=  iifiYzs' 

0.0273/  +JO.93436 

=  o.93^7&  /ae''i9'33" 

£&  129 
/  1"  40'  27" 

0.3  64  17  +J0.070Oi> 
=  0.370BSl/0''£3-'Zi' 

£3.SS 
\90°3Z'33" 

+1.98 

ZOO 

lOO 

0.0  8SS8+J/.99S-t(, 
0^=  II4°I9'£4" 

0.O3J43+J0.9/4SZ 
=■  0.9 1 S  ZZ  1  '&7°4h'  £-y 

£4  9£S 
/  2°  13'  07" 

0.41  3S4+J0.0783S 
=  0.42090  /I0''43'4  i" 

.60.77 
\90°42'/5'' 

¥2.iS 

Z25 

75 

O.O'jhkl   +J2.048S4 
e^-  II7'2Z'24" 

0.04449+ j  0.89218 
=0.89328  / 37° 08'4Z" 

£3  i>38 
/Z"  SI'  17" 

0.4 10194- +J0.0SS93 
^0.4h9  8i>  /lO'SZ'K," 

&7-8S 
\90''£3'40" 

4-3.40 

Z50 

SO 

0.I073S    +ds.lOI(>4 
©2=  l20°24S-i' 

O.0S44S+j0.Si,73S 
=  0.  S6  90S  /8&°24'28" 

£2/  83 
/3°36'32' 

0.SO9/7+JO.O9Z7S 

=  0.S/7£S  /icn'zi," 

,    7-5'.73 
\9/''0&'30" 

f  4-33 

27  S 

ZS 

O./ZSO?   i-j  2JS473 
02~  /23»27'22'' 

o.0i,SZS  +  J0.840  14 
=  O.Z4Z&7  /SS'SS'  3J" 

£0  599 
/4°  2  6' 2  7" 

0.SSSI4+J0.09874 
=  0.S6385  /lO'OS'OT' 

,      S/.^2 
\9l°20'49"  -^^■^' 

300 

o 

0./2  SSZ  +j  a.2079l 
02=   IZi?29'Si' 

O.076&3  +  JO.8/OS6 
=  0.8/4Z0  /84''36'0S" 

■48  889 
/  £'24'  ST 

0.£9973-i-jO. 103^4 
=  O.6  0  8i,S/9''4  9'Z2' 

S7.89          ,  ,  , 
\9l''3&'34"  +b.h4 

GRAPHS  OF  CURRENT  AND  VOLTAGE   (ZERO  LOAD  CONDITIONS) 

<a  90 

UJ 
(£ 

S  80 

S 
■*  70 

-  lUO 

^--^, 

^ 

-90  2 
a. 

on    ^ 

' 

70    "* 

O   '" 

3 

<  60 

""^^ 

. 

VOLTAGE 

-60    < 

< 
f  50 

" 

^ 

^^^ 

ks2£iyr 

-50    2 

3 

Z  *" 

Pin 

0 

^  20 

^'^^ 

> 

SENDING 
END 


50 


100  125  150  175  200 

DISTANCE  FROM  SENDING-END  (MILES) 


225 


250 


276  300 

RECEIVING 
END 


SINH(9|+je2)=(SINHe|  COS  92+ j  COSH  e,  SIN  82). 

COSH  (e|+je2)=(C0SH  e,  cos  Oj+j  sinh  e,  sin  e^). 

ONE  QUADRANT=  1 .67079632  CIRCULAR  RADIANS 
ONE  CIRCULAR  RADIAN=206264,B062"=57'  I  T  44.8" 


OC  =0.00042939+j  0.002 1  2336 


ANGLEAT  RECEIVING  END  0Ro=  0     +j  1.57080 

ANGLE  OF  LINE    0        =0. 1  2B82  +  J  0.63701 

eso=0+0Ro^°'^^^^''j^'^°^®' 


FIG.    50 — CURRENT  AND  VOLTAGE  DISTRIBUTION 

For  pri)I:/cni  X  by  position  angles  (zero  load  conditions). 


HYPERBOLIC  SOLUTION  FOR  LONG  LINES 


107 


PnoBLEM  "X" 

H  =  IOS    OHMS.        X=24<?0HMS. 
&=0«M0.  Y=  0.00/563  MHO. 

Z  =  2  70.2  33 /^7'Og'og* 
Y=  0.00 /S  6 3 190* 
6  =  -YzT  =^0.422  374  ^//^T'Qg'Og' 
=  0.64"?  903  J  /7^'3-t'04' 
=  0./2S8/6S-fj0.6  3  7  0092 


CALCULATION  FOR  *"*"  9 

e 

S//VH  e=0-l03  S393+J O.S9973S 
=  0.60g^.yg3/g0V0'3g"    * 

SiNHd _   O.COS  iS,S«  3  /80°/0'37' 
0  0.(3^9  903  S/7S'  34'04' 

=  0.93i53(,  S//''3i'33" 
=    It^PEZinNCE    CORRECTINC  FACTOR. 


CALCULATION  rOK    TAWHCB/a) 

-2=0.324  95/  g/Zg'S-fO-^" 
=  0.O644084  +  jO.3(  g  504  6 

■S/NH-j-=o.O*/2//  2  2  -i- J  0-3 13  796  3 
=  0.3/f  7/0  7/7g'37V3'' 

COSH-^=0.9S/  iVS'H-jO.OZO  /83  2 

fi^  SINH(0/2)   o^3,g  TIP  7/79*5743' 


>^**^  Z       CQSH(e/2)    0-9S/  8S94/I'  /2>S4* 

■  -=   Q.-33S  869  6  i72l±:il±±' 
TfiNH(6/z)_    0.33S  i(>9(>/77°44'49' 


e/Z  0.3X4  'fS/  8/78°  3^'0-r 

=  1.033  S9  8  \0''*'I'/S' 


9  6 

CHECK,  SlNH6  =  k  SINH-^COZHJ  =  2X.03I97/07  17  S'sY^^' X  Q.9S  I  SS9t- / 1°  I2's^ 

-  0.608  (iS8-^ / 80°/o'38'bNnicH  cM£CKi  with  *J. 
PIG.   48 — MATHEMATICAL   DETERMINATION   OF   CORRECTING   FACTORS    FOR   EQUIVALENT  IT   SOLUTION 


PROBLEM  "X" 


Z=/OS  +  ^'Z-4f=Z70-Z'33/67°08'0B'0HMS. 
Y=     O  +d0.O0ISi>3==O.O0ISi>3\90°  MHO. 

6=  fzY  =  o./zs8U?+jo.<>37oo9Z  Hyp. 

KV-flf,;^  -6000  00 oV25°  50'3  /'  WMTTS. 

=  S  400  00  0~j  2  IdI S  340. 
Ef^l^=  ho  044.4  VOLTS  TO  NEUTRHL. 
lf^=99.9z60S  \2S°  50'3I" 


SOLUTION  FOB  TANH  0. 


8  =  ^  =&00.888/2S'S0'3l'  OHMS. 

^-0  =  ^^=4  I  5. ■80S  \ll'Z5'S(i,"   OHMS. 
-r«MM  f)    -    ^00-SZ8lx£°  -SO'^'" 

=  I.44S/Z  J37'  Ko'27' 
:^l.l499S+3  O.S7S2Q9  * 

(Bii-dez) 


SOLUTION  FOR  ANGLE  0- 


TflNH-^{d,±j0,)  =  iLOGH\UL±%±^ 


{i-0,f+el 


+  J 


Tl-TRt^ 


:-l/^l+l 


iM 


+  THN' 


01-1 
±02 


f^    -2  1/  (/- 1.14  99S)Z  +o.375Z09^  ^  2 


--^  LOGH 


4:&2229+0.7&599  .     0  80''-67''  So'SS'4-9''43'zO') 

<?.02  2-f  8i5+0.76«?y        ^  a 


=  X('-OGH  Z.i,/4/5)  +  ji,0''S6'//'' 
=j{o.96o^39)+j/.063S397 

0^=  0.4-8  04i>9S+jl.06  3S3  97 
0  =  O.  I  Z9  8/  68-^dO-i>37009Z 
fis=  O.i>0  9  ZSk^4-jl.700S4  89 


CHECK 

SINH  0i^=/.OO6S7Z  /7(>'03'34J' 

COSH  e^  =0.i>9i,S3/38''^7'0?' 

TMNH  6    =    ^'^"^'^    =     l-OOi>S7Z  /7(,'OZ'3(,' 


COSH  9, 


0.(>9(>S3  / 3  S'>'^7' 0  9" 


—  1.14  9  9S  4-;}  O.S7'SZ09  (w«/c«  CHECKS  vvit»*). 


FIG.  49— POSITION  ANGLE  Ob  AT  RECEIVING-END 

Mathematical  determination  at  load  conditions. 


io8 


HYPERBOLIC  SOLUTION   FOR   LONG   LINES 


CHOICE    OF    VARIOUS    METHODS 

Two  graphical  and  two  mathematical  forms  of  so- 
lution for  circuits  of  long  electrical  length  have  been 
described  thus  far.  These  four  methods  have  been 
given  for  the  purpose  of  providing  a  choice  of  proce- 
dure for  the  beginner.  Graphical  solutions  are  more 
simple  and  more  readily  performed  than  mathematical 
solutions  and,  if  used  correctly  and  made  to  a  large 
scale,  will  yield  results  well  within  the  limits  of  per- 
missible error  for  power  transmission  circuits.  There 
is  always  a  possibility  of  error  with  any  method,  even 
though  the  solution  is  carefully  checked.  For  this  rea- 
son it  is  desirable  that  errors  be  guarded  against  by  the 
use  of  two  different  forms  of  solution.     For  instance 


f  SENDING 


wave,  then  it  will  be  necessary  to  take  their  effect  into 
account,  if  high  accuracy  is  essential.  In  such  a  case 
there  is  an  independent  solution  required  of  potential 
and  current  ^pr  pc^ch  single  frequency  in  turn,  as  though 
the  others  did  not  exist,  and  then  the  r.m.s.  value  at 
any  point  on  the  line  is  the  perpendicular  sum  of  the 
separate  frequency  values. 

A  detail  discussion  of  the  manner  of  including  the 
effect  of  harmonic  components  in  the  current  and  volt- 
age waves  is  quoted  below  from  Dr.  Kennelly's  "Artifi- 
cial Electric  Lines." 

"The  ordinary  complex  harmonic  impressed  e.m.f.  contains 
a  fundamental  frequency  associated  with  multiple  frequency 
harmonics.  The  nth  multiple  of  the  frequency  is  called  the  nth 
harmonic.  The  fundamental  may  thus  be  included  as  the  first 
harmonic. 

"In  order  to  deal  with  the  plural-frequency  case  quantita- 
tively, it  is  necessary  to  analyze  the  impressed  potential  wave 
iiito  its  harmonic  components.  As  is  well  known,  the  complete 
Fourier  analysis  of  a  complex  wave  may  be  written 


SENDINQ-ENO 


60  MILES 


-END 


GRAPH   OF  CURRENT 
(LOAD  CONDITIONS) 


•  RECEIVING-END 
FIG.    52 — POLAR    DIAGRAM    OF    CURRF.NT    AND    VOLTAGE   DISTRIBUTION    FOR    PROBLEM  X 


the  first  solution  could  be  made  by  making  use  of  the 
Wilkinson  charts  followed  by  its  accompanying  graphi- 
cal solution.  The  second  solution  could  then  be  made 
by  means  of  Dr.  Kennelly's  ratio  charts  XVIII  to  XXI, 
followed  by  its  accompanying  graphical  solution.  These 
two  methods  would  then  yield  results  obtained  by  two 
entirely  different  routes  and  methods  of  procedure. 
The  use  of  two  such  methods  would  constitute  check 
against  errors  being  made  in  either  solution. 

EFFECT  OF   HARMONIC  CURRENTS  AND  VOLTAGES 

The  foregoing  discussion  is  based  upon  the  assump- 
tion that  the  fundamental  wave  is  of  sine  shape  and 
consequently  free  from  harmonics.  If  harmonics  of 
considerable  magnitude  are  present  in  the  fundamental 


F,  +  V'i  sin  ut  +  V'i  sin  2«/  +  V,  sin  3"/  +  V'4  sin  4w<  -h 
-I-  F",  COS  w/  -f  V'\  cos  zw/  -I-  V",  cos  zot  -i- 

F"*  cos  4w<  + v°'*^     ('^ 

where  Fo  is  a  continuous  potential,  such  as  might  be  developed 
by  a  storage  battery,  ordinarily  absent  in  an  a.  c.  generator 
wave,  F'l,  F"i,  F'2,  V'\,  etc.,  maximum  cyclic  amplitudes  of 
the  various  sine  and  cosine  components.  The  even  harmonics- 
are  ordinarily  negligible  in  an  a.  c.  generator  wave;  so  that 
F's,  V"2,  V\,  V",,  etc.,  are  ordinarily  all  zeros.  If  we  count 
time  from  some  moment  when  the  fundamental  component 
passes  through  zero  in  the  positive  direction,  V"i  =  o  and  the 
series  becomes 

F'l  sin  at  -f-  V'l  sin  3<.)i  +  V\  sin  5w<  +  

F",  cos  3w<  +  V'\cos.  s<^t  +  .volts     (2) 

Compounding  sine  and  cosine  harmonic  components  into  result- 
ant harmonics  of  displaced  phase,  this  may  be  expressed  as 
Fr,  sin  at  +  Vn  sin  (aw*  -|-  /9.°)   +  Fr.  sin  (S«'  +  P»  )   + 
volts     (3) 

volts     (4) 


where 

and 


V.,  =  v"V\'  +  FV 

tan  ^n     =    yl 


numeric     (5) 


'  HYPERBOLIC  SOLUTION  FOR  LONG  LINES 


Formulas  (l)  and  (2)  give  the  wave  analysis  in  sine  and 
cosine  harmonics,  while  (3)  gives  it  in  resultant  sine  harmonics. 

"When  considering  a  plural -frequency  alternating-current 
line,  we  require  to  know  the  harmonic  analysis  of  the  impressed 
potential,  either  in  sine  and  cosine  harmonics,  or  in  resultant 
harmonics,  the  latter  analysis  is  preferable,  as  being  shorter 
and  containing  fewer  terms.  A  decision  must  be  made  as  to 
the  number  of  frequencies  or  upper  harmonics  which  must  be 
taken  into  account. 

"Ordinarily,  the  sizes  of  the  harmonics  diminish  as  their 
order  increases;  but  there  are  numerous  exceptions  to  this  rule, 
as  when  some  particular  tooth  frequency  in  the  alternatipg- 
currciit  generator  establishes  a  prominent  size  for  that  har- 
monic. Care  must  therefore  be  exercised  not  to  exclude  any 
important  harmonics.  On  the  other  hand,  the  fewer  the  har- 
monics to  be  dealt  with,  the  better,  because  the  labor  involved 
in  correctly  solving  the  problem  increases  in  nearly  the  same 
ratio  as  the  number  of  harmonics  retained. 

"The  rule  is  to  work  out  the  position  angle,  r.m.s.  potential, 
and  r.m.s.  current  distributions,  over  the  artificial  or  conjugate 
smooth  line,  for  each  harmonic  component  in  turn,  as  though 
it  existed  alone,  and  then  to  combine  them,  at  each  position,  in 
the  well-known  way  for  root  mean  squares. 

"Combination  of  Components  of  Different  Frequencies  into 
a  R.m.s.  Resultant. — Let  the  r.m.s.  value  of  each  alternating- 
current  harmonic  component  be  obtained  by  dividing  its  ampli- 
tude withi/^  in  the  usual  way,  and  let 

y.  =— ir  =  ^r  "  +'^  °  r.m.s.  voJts    (6) 

be  the  r.m.s.  value  of  the  «th  harmonic.  Then  the  r.m.s.  value 
of  all  the  harmonics  together,  over  any  considerable  number  of 
cycles,  will  be 

V  —  V  TvTTTTTT  +  r.ni.s.  volts     (7) 

or,  as  is  well  known,  the  joint  r.m.s.  value  of  a  plurality  of 
r.m.s.  values  of  different  frequency,  is  the  square  root  of  the 
sum  of  their  squares.  If  a  continuous  potential  Fo  be  present, 
this  may  be  regarded  as  a  r.m.s.  harmonic  of  zero  frequency, 
and  be  included  thus  : 

V  —  V  Vo  +  Vi  +  V'  -f  F,'  +  . .  .r.m.s.  volts  (8) 
Moreover,  from  (4),  it  is  evident  that  the  squares  of  the  r.m  •>. 
values  of  the  sine  and  cosine  terms  of  any  harmonic  may  be 


FIG.   53 — GEOMETRICAL    REPRESENTATION    OF    A    JOINT    R.M.S.    VALUE 

OF   PLURAL-FREQUENCY   COMPONENTS    DY    PERPENDICULAR 

SUMMATION    OR    "CRAB    ADDITION" 

substituted  for  the  square  of  their  resultant;  or  that,  in  this 
respect,  the  sine  and  cosine  terms  may  be  treated  as  though 
they  were  components  of  different  frequencies. 

"The  same  procedure  applies  to  plural-frequency  currents. 
Find  the  r.m.s.  resultant  harmonics.  The  r.m.s.  value  of  all 
together  will  be  the  square  root  of  the  sum  of  their  squares.  A 
continuous  current,  if  present,  may  be  included,  as  the  r.m.s. 
value  of  an  alternating  current  of  zero  frequency. 

"Graphical  Representation  of  R.m.s.  Plural-frequency  Com- 
bination.— The  process  represented  algebraically  in  (7)  or  (8) 
may  be  represented  eraphirally  by  the  process  of  successive  pf  >■ 
pendicular  summation,  or  "crab  addition."  An  example  will 
suffice  to  make  this  clear.  A  fundamental  alternating  current 
of  100  amp.  r.m.s.,  is  associated  with  a  continuous  current  of 
50  amp.,  and  with  two  other  alternating  currents  of  other  fre- 
quencies of  20  and  10  amp.  r.m.s.,  respectively.  What  will  be 
the  joint  r.m.s.  current?     Here  by  (8),  », 

/  =\/  100'  -f-  50'  +  20'  -f  10'  =  V  10 000  -f  2500  -f  400  +  100 
=  1/  13000  ^  114.0175  amp.  r.m.s. 
"In  Fig.  53,  OA  represents  the  fundamental  r.m.s.  current. 
AB,  added  perpendicularly  to  OA  represents  the  continuous 
current,  or  current  of  50  r.m.s.  amp.  at  zero  frequency.  The 
perpendicular  sum  of  OA  and  AB  is  OB  =  1 11.8034  amp. 
Adding  similarly  the  other  frequency  components  BC  and  CD, 


109 

the  total  perpendicular  sum  is  OD  =  114.0175  amp.  The  order 
in  which  the  components  are  added  manifestly  does  not  affect 
the  final  result,  and  it  is  a  matter  of  insignificance  whether  the 
various  frequencies  coacting  are  "harmonic,"  t.  e.,  are  integral 
multiples  of  a  fundamental,  or  not,  so  long  as  they  are  different. 
"The  complete  solution  of  an  alternating-current  line  with 
complex  harmonic  potentials  and  currents  thus  requires  an  in- 
dependent solution  of  potential  and  current  for  each  single 
frequency  in  turn,  as  though  the  others  were  non-existent,  and 
then  the  r.^n.s.  value  at  any  point  on  the  line  is  the  perpendicular 
sum  of  the  separate  frequency  values.  The  powers  and  energies 
of  the  different  frequencies  are  independent  of  each  other,  and 
the  total  transmitted  energy  is  the  sum  of  the  energies  trans- 
mitted at  the  separate  component  frequencies. 

BIBLIOGRAPHY 

In  order  to  give  due  prominence  to  some  of  the  valuable 
contributions  on  the  subject  of  performance  of  electrical  circuits 
and  as  an  acknowledgment  to  their  authors  of  the  assistance 
received  from  a  study  of  them,  the  following  publications  arc 
suggested  as  representing  a  very  helpful  and  valuable  addition 
to  the  library  of  the  transmission  engineer.  They  are  given  in 
the  approximate  order  of  their  publication: — 

Calculation  of  the  High  Tension  Line  and  Output  and 
Regulation  in  Long  Distance  Lines  by  Percy  H.  Thomas. 
(Published  in  A.  L  of  E.  E.  Trans.  Vol.  XXVIII,  Part,  I, 
1909).  The  former  paper  introduces  a  so-called  "wave  for- 
mula" for  determining  the  performance  of  long  lines  having 
considerable  capacity  which  embodies  the  use  of  algebra  ofoXy. 
The  second  paper  suggests  the  use  of  split  conductors  in  order 
to  adjust  the  ratio  of  the  capacity  and  inductance  of  the  line 
so  that  the  leading  and  lagging  components  more  nearly 
neutralize  each  other. 

Formulae,  Constants  and  Hyperbolic  Constants  by  W.  E. 
Miller.  (Published  in  G.  E.  Review,  supplement  dated  May 
1910).  This  is  a  treatise;  upon  the  subject  wherein  hyperbolig 
functions  of  complex  angles  are  tabulated  for  sinh  and  cosh 
(x  +  jy)  up  to  X  :=  I,  y=:  I  in  steps  of  0.02. 

Transmission  Line  formulas  by  H.  B.  Dwight  (Published 
by  John  Wiley  &  Sons,  Inc.).  This  book  introduces  what  are 
known  as  "Dwight's  'K'  formulas,"  which  permit  the  solution 
of  transmission  problems  without  the  use  of  mathematics 
higher  than  arithmetic.  It  also  contains  working  formulas 
based  upon  convergent  series  and  the  solution  of  many  problems 
both  by  the  K  formulas  and  by  convergent  series. 

Tables  of  Complex  Hyperbolic  and  Circular  Functions  by 
Dr.  A.  E.  Kennelly.  (Published  by  the  Harvard  University 
Press).  This  book  gives  functions  of  complex  angles  for  polar 
values  up  to  3.0  by  steps  of  o.i  and  for  angles  from  45°  to  90" 
by  steps  of  one  degree;  also  functions  in  terms  of  reactangidar 
coordinates  x  -f-  jy  to  x  =  10  by  steps  of  0.05  and  of  y 
virtually  to  infinity  by  steps  of  0.05. 

Chart  Atlas  of  Complex  Hyperbolic  and  Circular  Functions 
by  Dr.  A.  E.  Kennelly.  (Published  by  Harvard  University 
Press  in  large  charts,  48  by  48  cm.)  Presenting  curves  for  M 
the  tables  published  in  above  referred  to  "Tables  of  Complex 
Hyperbolic  and  Circular  Functions"  for  rapid  graphical  inter- 
polation.    J 

Constant  Voltage  Transmission  by  H.  B.  Dwight.  (Pub- 
lished by  John  Wiley  &  Son,  Inc.).  Embraces  a  very  complete 
study  of  the  use  of  over-excited  synchronous  motors  for  con- 
trolling the  voltage  of  transmission. 

The  Application  of  Hyperbolic  Functions  to  Electrical 
Engineering  Problems  by  Dr.  A.  E.  Kennelly.  (Published  by 
the  McGraw-Hill  Book  Company).  Every  student  should  have 
a  copy  of  this  book  because  of  its  simplicity  and  completeness 
in  explaining  the  application  of  hyperbolic  functions  to  trans- 
mission circuit  problems.  It  also  contains  a  very  complete 
bibliography  of  publications  upon  this  general  subject. 

Artificial  Electric  Lines  by  Dr.  A.  E.  Kennelly.  (Published 
by  McGraw-Hill  Book  Co.).  This  is  a  valuable  treatise  in 
which  the  subject  is  treated  in  accordance  with  the  hyperbolic 
theory.     , 

Electrical  Phenomena  in  Parallel  Conductors  by  Dr.  F.  E. 
Pernot.  (Published  by  John  Wiley  &  Son,  Inc.).  Being  a 
very  recent  treatise,  this  book  contains  much  practical  and 
many  readily  understandable  explanations  for  both  the  beginner 
and  those  further  advanced  in  the  study  of  this  subject.  It 
contains  a  six-place  table  of  logarithms  of  real  hyperbolic 
functions  for  values  of  .r  from  0.000  to  2.000  for  intervals  of 
o.ooi  in  the  argument.  This  is  the  most  complete  table  of  real 
hyperbolic  functions  which  the  author  has  seen. 


tio 


HYPERBOLIC  SOLUTION  FOR  LONG  LINES 


FABLE  P— SUBDIVISIONS  OF  A  DEGREE 


SECONDS 

TO 
DEGRE-:S 

MINUTES 

TO 
DEGREES 

DEGREES 

TO 

MINUTES  AND  SECONDS 

1!=       o 

!=       o 

°=          f            ff 

°=        f        If 

01 
02 
03 

0.OOO3 
O.OOOh 

o.oooe 

Ol 

oz 

03 

O.OIbJ 
0.0333 
0.  0  500 

0.00/ 
0.002 
0.00  3 

00      03J, 
00      07.2 

00   /o.e 

0.006 
0.007 
0.008 

00     2/.6 
00     25.2 
0  0     2S.8 

04 
06 

O.OOI  1 
0.OO/4 
O.OOI7 

04 
03 

06 

O.Okh7 
0.O833 
O.IOOO 

0.004 

o.oos 

00     14.4 
00     18.0 

0.00  9 
0.0  10 

00     32.4 
00     3t.O 

07 
OS 
09 

0.OOI9 

o.  oozz 
o.oozs 

07 
08 
09 

O.llhl 
O.I  333 
O.I  SOO 

/O 
// 

O.0O28 
0.OO3I 
0.0033 

10 
II 
12 

0.li>i,7 
0./833 
O.2000 

0.01 
o.oz 

00       36 
0/         /2 

O.SI 

o.sz 

30      36 

3/       IZ 

/3 
IS 

O.OOSi 
O.0039 
0.00-f2 

13 
14 
IS 

0.2/67 
0.2333 
O.2500 

0.03 
0.04 

o.os 

01       48 
OZ       24 
03       00 

0.33 

O.S4 
O.SS 

3/      48 
3i    24 
33      00 

lb 
n 

IS 

0.00-+4 
0.0047 
0.00.50 

Ih 
17 
18 

0.2667 
0.2833 
0.3000 

O.Oh 
0.07 
0.08 

03  36 

04  IZ 
04     48 

O.Sh 
0.57 
O.SS 

33  36 

34  IZ 
34     48 

/<? 

ZO 
Zl 

0.0053- 
0.0  055 
0.O0.S8 

19 
ZO 
Zl 

0.3/67 
0,3333 
0.3SOO 

0.09 
0.10 
0.  II 

05  Z4 

06  00 
06      36 

O.S9 
O.hO 
O.hl 

35  24 

36  00 
36      36 

22 
23 

24 

O.OOi/ 

o:ooi4- 

0-00,7 

22 
23 

24 

0.3667 
0.3833 
0.4000 

O.IZ 
O.I3 
0.14 

07      IZ 

07  48 

08  Z4 

0.62 
0.63 
0.64 

37       IZ 

37  48 

38  24 

■  ZS 

ze, 

27 

O.OOh9 
0.007Z 
0.0073 

ZS 

Zh 
27 

0.4/67 
0.4  333 
O.4S0O 

O.IS 

0.1  h 

0.1  7 

09     00 
09     36 
/o      /z 

0.6S 
0.66 
0.67 

39      00 

39  3h 

40  IZ 

28 

2<? 
30 

■  0.0078 

o.ooai 

O.OO  83 

is 

29 
30 

a4667 
0.4833 
0.3-000 

0.18 
0.1  9 
0.20 

10  48 

11  Z4 
IZ     00 

0.68 
0.69 
O.70 

40     48 
■4  1      24 
4Z      00 

3/ 
32 
33 

0.0086 
0.0089 
O.O09Z 

31 
32 
33 

O.Sib7 
0.333'3 
0.3S00 

0.2/ 
0.22 
0.23 

IZ     36 
/3      /Z 
/3     48 

0.7 1 

0.7  z 

0.73 

42  36 

43  /2 
43       48 

34 
35 
3t 

0.0O9^ 
0.0097 

o.oioo 

34 
3S 
34 

0.3  667 
0.3833 
0.  6000 

0.24 
0.23 
b.  2  6 

14  Z4 

15  00 
/S    3t> 

0.74- 
0.7.5 
0.76 

44  Z4 

45  00 
4S      36 

37 
39 

0.0I03 
0.01  of, 

o.oioe 

37 
38 
39 

0.6/67 
0.6333 
0.6300 

0.27 
0.2s 
0.29 

If,      IZ 
/6     48 
17     Z4 

0.77 

0.7  S 

0.79 

46  IZ 
4l>     48 

47  24 

-*< 
■42 

O.O  II  1 
0.0//4 
O.O/  17 

4-0 
.41 
■4Z 

0.6667 
0.6  833 
0.7000 

0.3O 
0.3/ 
0.32 

18     00 
18     36 
/9      /2 

O.80 
0.8I 
0.82 

48      00 

48  36 

49  IZ 

43 
44 
■4S 

0.0119 
O.OIZZ 

o.oizs 

43 

44 
45 

0.7/67 
Q.7333 
0.  7.J0O 

0.33 
0.34 
0.35 

19    48 
ZO    24 
Zl    00 

0.  83 
0.84 
O.SS 

49  48 

50  24 

51  00 

46 
47 
48 

O.O  128 
0.OI30 
0.0/33 

46 
47 
48 

0.7hh7 
0.7833 
0.8000 

0.36. 
0.37 
0.38 

Zl     36 
22     /2 
22    48 

0.86 
0.87 
O.SS 

SI     36 
32      IZ 
SZ     48 

.4? 
30 
J/ 

O.O  13b 

o.to;3? 
O.O  14  1 

49 
SO. 
SI 

0.8/67 
0.8333 
0.  &30O 

0.3  9 

0.4:0 

■  0.41 

■  23      24 
24      00 
24     36 

0.89 
0.90 
0.9/ 

53  24 
34     00 

54  36 

.>y2 

.53 
54 

O.OJ44 

•0.0)47 

O.OISO 

SZ 
33 
S4 

.0,86  6' 7 
0.8833 
0.9000 

0.4Z 
0.43 
0.44 

ZS     /2 
2S    48 
2  6      24 

0.9Z 
0.93 
0.94 

SS      IZ 

55  48 

56  34 

^5 
Si 
,S7 

O.OIS3 
O.0IS6 
O.OIS9 

SS 
Si, 
37 

0.9 167 
0.9333 
0.9SOO 

0.4S 
0.46 
0.47 

27  00 
Z7      36 

28  /2 

0.9  S 

0.96 
0  .97 

£7     00 

57  36 

58  IZ 

.5? 
60 

O.OIhZ 
o.oH,4 

o.oihr^ 

38 
S9 
60 

0.9667 
0.9  833 

1 .0000 

0.48 
0.49 
O.SO 

2S    48 

29  S4 

30  00 

0.93 
0.99 

/•oo 

58  48 

59  S4 

60  00 

0.^1  = 
o\oos  = 


CfZ4'3h'^ 

o'oo'  la* 


(f4i'od'- 

Cf0<f4tl'  - 


0'.i>933. 
O'.OIZB. 


CHAPTER  XII 
COMPARISON  OF  VARIOUS  METHODS 


The  "localized  capacitance"  or  "localized  admittance"  methods  are  discussed  below  lor  the  t\vo  foUowmg 
reasons.  A  discussion  of  them  is  of  academic  i^terc^t  and  a  tabulation  of  the  magnitude  of  the  errors 
in  the  results  as  obtained  by  these  approximate  methods  when  applied  to  circuits  of  different  lengths  and 
frequencies  should  be  helpful.  These  methods  may  be  carried  out  either  graphically  or  mathematically,  but 
since  thoy  are  only  approximate  the  simpler  graphical  solution  should  suffice.  Their  principle  virtue  is 
the  fact  that  they  simplify  the  determination  of  performance,  but  this  is  obtained  at  the  expense  of  accuracy. 
The  more  accurate  of  these  methods  is  somewhat  tedious  to  carry  out.  The  graphical  solution  previously 
described  in  connection  with  the  Wilkinson  charts  will  be  generally  more  accurate  and  shorter  than  these 
localized  capacitance  methods. 


T 


HE  LOCALIZED  CAPACITANCE  methods 
are: — the  single  end  condenser  method;  the 
middle  condenser  or  T  method;  the  split  con- 
denser or  nominal  ir  method  and  Dr.  Steinmetz  three 
condenser  method.  These  four  lumped  capacitance 
methods  assume  the  total  capacitance  of  the  circuit  as 
being  divided  up  and  "lumped"  in  the  form  of  conden- 
sers shunted  across  the  circuit  at  one  or  more  points. 

PROBLEM  "X" 

KV-Aln~  '•°°°  "^v*         Ern"  80-0**  W\.T^ 

KWln"    ''*00  XW  pP|^- 90%  UQQINO 

ll_  -  99.91  AMPERES         F  -  SO  CYCLES 
RECEIVING 
END 

98.92  AMPS, 


methods,  usually  an  approximation  to  the  true  value  may 
be  obtained. 

The  middle  condenser  or  T  method  assumes  that 
the  total  capacitance  may  be  shunted  across  the  circuit 
at  the  middle  point.  On  this  assumption  the  total 
charging  current  will  flow  over  one  half  the  length  of 
the  circuit.  This  method  is  therefore  more  nearly  ac- 
curate than  the  single-condenser  method. 

The  split  condenser  or  ir  method  assumes  one  half 


LINEAR  CONSTANTS 
R  -  106  OHMS 
X  -  249  OHMS 
B  -0.001663  MHO 
Q  -  0  (80  THAT  Y-B) 


•g>3'l8'  71'  PF^9».9a7*LEAOINO 


NEUTRAL 

RESULTS  CALCULATED  BY 

RIGOROUS  SOLUTION 

SINGLE  END 

CONDENSER 

METHOD 

%  ERROR 

EsN-  70,662  VOLTS 

63,329  VOLTS 

-  10.37% 

Ig-  94.75  AMPS. 

103.038  AMPS 

+    8.8% 

PFg-  ♦  93.42% 

+  99.927% 

+    7,0% 

LOSSn-865KW 

1120  KW 

+  30.9% 

fR-lOJI»VOLT») 


93.862  AMPERES 


FIG.    54 — SINGLE  END  CONDENSER   METHOD 

Problem  X. 


The  single  condenser  method  assumes  the  total  ca- 
pacitance as  being  lumped  or  shunted  across  the  circuit 
at  the  receiving-end.  On  this  assumption  the  total 
charging  current  for  the  circuit  would  flow  over  the 
entire  circuit.  Actually  the  charging  current  is  dis- 
tributed along  the  circuit  so  that  the  entire  charging 
current  does  not  flow  over  the  entire  circuit.  Obviously 
the  assumption  of  the  total  capacitance  being  lumped 
at  the  receiving-end  will  therefore  give  over  compensa- 
tion for  the  effect  of  the  charging  current  upon  the 
voltage  regulation  of  the  circuit.  This  method  of  solu- 
tion yields  a  voltage  too  low  at  the  sending  end  by 
nearly  the  same  amount  that  the  straight  impedance 
method  gives  it  too  high.  By  averaging  the  values,  as 
obtained  by  the  impedance  and  single  end   condenser 


the  capacitance  being  shunted  across  the  circuit  at  each 
end.  In  this  case  one-half  of  the  charging  current  .flows 
over  the  entire  circuit.  This  assumed  distribution  of  the 
charging  current  also  more  nearly  represents  the  actual 
distribution  than  the  single-condenser  method. 

Dr.  Steinmetz  has  proposed  a  method  assuming 
three  condensers  shunted  across  the  circuit.  One  in  the 
middle,  of  two-thirds,  and  one  at  each  end,  each  of  one 
sixth  the  total  capacitance  of  the  circuit.  This  method 
is  equivalent  to  assuming  that  the  electrical  quantities 
are  distributed  along  the  circuit  in  a  way  representing 
ap  arc  of  a  parabola.  This  method  assumes  one-sixth 
the  charging  current  flowing  over  one  half  the  entire 
circuit  and  five  sixth  the  charging  current  flowing  over 
the  other  half  of  the  circuit.     This  method  gives  quite 


112 


COMPARISON  OF  VARIOUS  METHODS 


accurate  results  unless  the  circuit  is  very  long  and  the 
fiequency  high. 

Figs.  54-57  show  leaky  condensers  placed  at  dif- 
ferent points  of  the  circuits,  that  is  they  indicate  that 
there  is  a  leak  G,  as  well  as  a  susceptance  B.  For  sim- 
plicity pure  condensers  have  been  assumed  in  the  ac- 
companying calculations ;  that  is  we  have  assumed  G=o. 
This  is  the  usual  assumption  in  such  cases,  for  the 
reason  that  G  is  usually  very  small,  and  localized  ca- 
pacitance methods  are  approximations  at  best.  In  the 
equivalent  w  solution  previously  given,  we  have  in- 
dicated the  treatment  when  the  condensers  have  a  leak. 
In  such  case,  however,  the  equivalent  ir  method  pro- 
duces exact  results,  and  the  nature  of  such  solution  may 
demand  a  condenser  having  a  material  leak. 

AUXILIARY  CONSTANTS 

Mr.  T.  A.  Wilkinson  and  Dr.  Kennelly  have 
worked  out  the  algebraic  expressions  for  the  auxiliary 

PROBLEM  "X' 


receiving-end.     In  such  case  the  entire  charging  current 
would  flow  over  the  total  length  of  the  circuit. 

Solution  by  Impedance  Method — The  diagrams  of 
connections  and  corresponding  graphical  vector  solution 
for  problem  X  by  the  single  end  condenser  method  is  in- 
dicated by  Fig.  54.  The  current  DN  consumed  by  the 
condenser  (zero  leakage  assumed)  leads  the  receiving- 
end  voltage  OR  by  90  degrees  and  is, — 

/c  =  0.CX)I563  X  60.046  =  93.852  amperes. 

The  load  current  of  99.92  amperes,  lagging 
25°  50'  30"  (90%  power-factor)  has  a  component  OA 
of  pp.p^  X  o.po^=  89.928  amperes  in  phase  with  the  re- 
ceiving-end voltage  and  a  component  AD  of  99.92  X 
^■4359  =  43-555  amperes  in  lagging  quadrature  with  the 
receiving-end  voltage.  This  lagging  component  is ' 
therefore  in  opposite  direction  to  the  charging  current, 
the  effect  of  which  is  to  neutralize  an  equivalent  amount 
of  charging  current.  The  remaining  current  AN  in 
leading  quadrature  with   the  receiving-end  voltage   is 


KV-Aln-  8.000  kva 
KWln-  »-*ookw 

^-  89,02  AMPERES 


Erm- 

F- 


60.048  VOLTS 
80%  LAQOINQ 


R 


ee.siTAMPS. 


V 


X 

^00000- 


46.826  AMPar 


fl.SeiAMPS. 


LINEAR  CONSTANTS 

R  —  106  OHMS 
X  -  249  OHMS 
B-  0.001663  MHO 
Q-0  (80THATY-B) 


NEUTRAL 

RESULTS  CALCULATED  BY 

NOMINAL  SPLIT 

RIGOROUS  SOLUTION 

CONDENSER 
METHOD 

%  ERROR 

EgN- 70,662  VOLTS 

72,318  VOLTS 

*2.3t% 

Ig-  84.76  AMPS. 

81.862  AMPS. 

-2.84% 

PFg-  +  83.42% 

-•-83.878% 

-►0.68* 

LOSSn'  b^skw 

861  KW 

-047% 

IR-  8.449  VOLTS 


60.046  VOLTS 


46.926  AMPERES 


FIG.    SS — NOMINAL    V   OR    SPLIT   CONDENSER    METHOD 

Problem  X. 


constants  corresponding  to  these  four  circuits  of  local- 
ized capacitance.  These  are  given  in  Table  Q.  It  may 
be  interesting  to  observe  to  what  extent  each  of  the  four 
localized  capacitance  methods  takes  account  of  the  three 
linear  line  constants  R,  X  and  B.  The  rigorous  or  exact 
expression  for  the  auxiliary  constants  is  given  under 
Table  Q  for  comparison  with  the  values  corresponding 
to  the  localized  condenser  methods.  The  numerals  un- 
der the  algebraic  expressions  correspond  to  problem  X ; 
that  is,  to  a  certain  60  cycle  circuit,  300  miles  long. 
They  are  given  to  illustrate  for  a  long  circuit,  the  ac- 
count taken  of  the  fundamental  constants  for  each  of 
the  five  methods  listed.  These  numerals  may  be  com- 
pared with  the  rigorous  or  exact  values  as  given  under 
the  rigorous  expressions  at  the  bottom  of  the  table. 

SINGLE  END  CONDENSER  METHOD 

This  method  assumes  that  the  total  capacitance  of 
the  circuit  may  be  concentrated  across  the  circuit  at  the 


93.852  —  43-555  =  50.297  amperes.     The  current  ON 

in  the  conductor  is  therefore: — 

/,  =  V  (89.928  '  -f  (50.297)' 
=  103.038  amperes. 

The  current  at  the  sending-end  leads  the  voltage;  at 
the  receiving-end  by  the  angle  ^r  whose  tangent  is, — 

50-297  „      ,    ,„ 

8^:5^  =  ^     '3'  06" 

The  voltage  consumed  by  the  resistance,  and  the  re- 
actance of  each  conductor  is, — 

IR  =  103.038  X  105  =  10 819  Volts  (resistance  drop) 
IX  =  103.038  X  249  =  25656  Volts  (reactance  drop) 

The  receiving-end  conditions  are  thus, — 

/b  =  103.038  amperes 

Or  ==  29°  13'  06" 
Cos  en  —  0.8772 
Sin   Ob  =  0.4881 

and  from  (40)  *•- 


COMPARISON  or  VARIOUS  METHODS 


"3 


£«  = 


V.(60  046  X  0.8727  +  10  819)'  +  (60  046  X  0.4881  —  25  656)' 
=  63  329  \3°  18'  27"  volts  to  vector  ON 
=  63  329  /25°  54'  39"  voiti  to  vector  of  reference. 
PF,  =  Cos  /3°  18'  27"  =  99927  percent  leading. 
KV-A,»  =:  103.038  X  63.329  =  6525  kv-a. 
KU',«  =  6S25  X  0.99927  =  6520  kiv. 
Loss«  =  6520  —  5400  =1120  kw. 

Solution  by  Complex  Quantities — From  Table  Q 
the  auxiliary  constants  corresponding  to  the  single  end 
condenser  method  are  found  as  follows: — 

fli  =  I  —  XB  =  0.610813 

(h  =  RB  =  0.164115 

6j  =:  /?  =  105  ohms. 

bi  =  X  =^  249  ohms. 

c,  =  0 

c,  =  B  =  0.001563  mho. 

The  voltage  at  the  sending  end  is  determined  as 
follows : — 

Jv  {Cos  *L  —  /  Sin  Ol)  =  89.928  —  ;  43.555 

X  (b,  +  ;■  b,)  —20286  +  1  17819 

+£..  (o«  +  /  Os)  =  36677  4-  /    9854 


£..  =  56963  -f-  /  27673 


J 


end  is  completely  determined  by  the  load  current  at  the 
receiving-end  and  the  vector  addition  thereto  of  the  cur- 
rent supplied  at  that  end  to  the  condenser  under  receiv- 
ing-end voltage.  For  determining  the  sending-end  volt- 
age A'y  =  I  -\-  YZ  and  B'y  =  Z;  but  for  determining 
the  sending-end  current  A'l  =  /  and  C'l  =  Y.    If  the 


condenser    were    applied 
would  be  identical. 


symmetrically    A'y  and  A'l 


=  63  329  725°  54'  39"  volts. 


SPLIT  CONDENSER  OR  NOMINAL  w  SOLUTION 

This  method  assumes  that  the  total  capacitance  of 
the  circuit  may  be  concentrated  at  the  two  ends,  one- 
half  being  placed  across  the  circuit  at  either  end.  In 
this  case  one-half  the  charging  current  flows  over  the 
entire  circuit.  The  total  resistance  and  the  total  react- 
ance of  one  conductor  is  placed  between  the  two  ter- 
minal condensers. 

With  this  assumption  the  current  consimied  by  the 
condenser  across  the  receiving-end  of  the  circtiit  is 
added  vectorially  to  the  load  current  and  the  power-fac- 
tor of  the  combined  currents  calculated.  With  these 
new  load  conditions  determined  the  conditions  at  the 


TABLE   0— AUXILIARY   CONSTANTS 
CORRESPONDING  TO  CIRCUITS  OF  LOCALIZED  CAPACITANCE 


Oa 


EQUIVAL£NT  OONVEROEHT  SCRIES 
FOAM  OF  EXPRESgKJW 


n 

-106 


A'-i 


B'-i 


C'-o 


8INQLE  END  OONDENSEft 


€ 


1  -  XB 

-o.aioeia 


RB 
•  ♦j0.te4M6 


R 
■I0> 


-♦jM» 


B 
♦  J0.00I683 


A'  -  I  ♦  »i     B' 


C- 


DOUBLE  END  OONOCN8EK 


(55 


XB 
9 


3 
•io.osio» 


R 
•lOB 


4 
- -0.000  0*41 


■-•10.001411 


A'-(.»?)  B'-z   C'-»(i*J?) 


MIDDLE  OONDEN8CA 


(~)  NOMINAL  "T  Bb 


J5 


-Wj  0.083069 


x-|(xa-R2) 
•  *i33e.06( 


B 
>jO.OOI6a9 


A'-(i+^)  B'-z('»^) 


C" 


THRCe    OONOENBCn 


(gift       \^   ~lfjg 


'-¥-i(x'-R») 


RB       RXB^ 

2     "       18 
'4  JO.O7S5O0I 


x-|Ua-R2) 

-  ►1236.721 


-*jH(x'-R») 


♦jO.OOUTM 


*  THE  eXACT  OR  RIOOROUB  tXPRCBSlONS  FOR  THE  AUXIUARV  00NSTANT8  ARE  OlVtN  BE  LOW  THE  NUMERICAL  FIOURE8  OORRES»ONO  TO  PROBLEM  "X" 

A  —  Vi+ir*   34  +  730  '  40  330*  '  B    -*i'*  8  *T3r*ro40*5iO»*  '  C   — ''('*T^+-i30"+6345*3e5l85*  ) 

-•OOeH8-0.8l068*j007883  — Z.8INHe-8>  74S8*)336880  —  T.BINHO- -0.000041 1«|  0.00I4884 


which  checks  exactly  with  the  results  as  obtained  pre- 
viously by  the  impedance  method. 

The  current  at  the  sending  end  may  be  determined 
as  follows: — 

It.  (Cos  0,.  —  /■  5"i»i  01.)  =   89928  —  ;  43-555 
-»-  £,.  (fi  -f- ;  f.)  =  o  +  ;  93852 


/•  =    89.928  -f-  ;'  50.297 

=  103.038  /29°  13'  06"  amperes. 

which  also  checks  exactly  with  the  result  as  previously 
determined  by  the  impedance  method. 

It  should  be  noted  here  that  in  determining  the 
sending-end  current,  the  auxiliary  constant  (a  -f-  ;  a, 
did  not  enter  into  the  calculation  as  it  does  in  the  rigor- 
ous solution ;  this  is  owing  to  the  inherent  dissymmetry 
of  the  single-end  condenser.  This  is  the  only  case  in 
which  the  capacitance  is  applied  dissymmetrically,  con- 
sequently the  current  entering  the  line  at  the  sending- 


*i« 

sending-end  are  calculated  by  the  impedance  method. 
I'his  is  the  only  calculation  required  when  employing 
the  nominal  w  method  for  determining  the  sending-end 
voltage.  The  voltage  at  the  sending-end  is  therefore 
more  readily  calculated  by  this  method  than  by  the  T 
method  which  requires  the  calculation  of  the  two  sepa- 
rate halves  of  the  circuit.  If,  however,  the  current, 
power-factor  and  kw  input  are  required,  a  second  calcu- 
lation must  be  made  to  determine  them.  In  such  cases 
the  current  consumed  by  the  condenser  at  the  sending- 
end  must  be  added  vectorially  to  that  of  the  line  conduc- 
tors. 

Solution  by  Impedance  Method — The  diagrams  of 
connections  and  corresponding  graphical  vector  solu- 
tions for  problem  X  by  the  nominal  w  method  is  in- 
dicated in  Fig.  55.  The  charging  current  consumsd  by 
the  condenser  (zero  leakage  assumed)  at  the  receiving- 


114 


COMPARISON   OF   VARIOUS   METHODS 


end  of  the  circuit  leads  the  receiving-end  voltage  by  90 
degrees  and  is, — 

0.001563 


/cr 


X  60046  =  46.926  amperes. 


The  current  /,  in  each  conductor  is  the  vector  sum 
cf  the  load  and  condenser  currents  and  may  be  deter- 
mined as  follows: — 

/r  =  1/  (9QC)2  X  0.90)'  +  (/cr  +  9992  X  -0.4359)' 
=  89.991  /2°  08'  a&"  amperes. 
PFr  =  Cos    2°  08'  48"  =  99.33  percent  leading. 

The  voltage  consumed  by  the  resistance,  and  the 

reactance  of  each  conductor  is, — 

IR  =  89.991  X  105  =     9449  volts    {resistance    drop) 
IX  =  89.991  X  249  =  22408  volts   (reactance  drop) 

and  from  (40), — 

V'(60046  X  0.9933  -f  9449)"  -f  (60046  X  0.037458—  22408)' 

=  72  JI9  /i6°  u'  08"  volts  to  current  Vector  OP. 

=  72319  /i8°  19'  .=i6"  volts  to  vector  of  reference  OR. 

The  charging  current  consumed  by  the  condenser 
at  the  sending-end  (zero  leakage  assumed)  leads  the 
voltage  at  the  sending-end  by  90°  and  is, — 

0.001563 
h,  = X  72319  =  56.517  amperes. 

The  current  at  the  sending-end  is  the  vector  sum 
of  the  current  in  the  conductor  and  the  current  con- 
sumed by  the  condenser  at  the  sending-end.  It  may  be 
calculated  as  follows: — 

OT  =  89.991   (.Cos  16°  II'  08")  =  86.424  amperes. 
TP  =  89.991   (Sin  16°   11'   08")   =  25.085  amperes. 
TN  =  56.517  —  25.085  =  31-432  amperes. 
therefore, — 


/.  =  1/86.424'  +  31-432-' 

=  91.962  /i9°  so'  07"  amperes  to  vector  OS. 
=  91.962  738°  19'  03"  to  vector  of  reference  OR. 
rr,  —  Cos  19°  59'  07"  =  93  979  percent  leading. 
KV-A..  =  91-962  X  72.319  =  6651  kv-a. 
KW.U  =  6651  X  0.93979  =  6251  kw. 

Loss,  ::=  6251  —  5400  =  851   kw. 

5400  X  100 


Eg   = 


6251 


=  86.37  percent. 


Solution  by  Complex  Quantities — From  Table  Q 
the  auxiliary  constants  corresponding  to  the  nominal  ir 
method  of  solution  are  found  as  follows : — 


XB 
at  =  1 :^—  0.8054065. 

a,  =  —^  =  0.0820575. 

61  =  R  =  105  ohms. 
bt  ■=  X  =:  -\-j  249  ohms. 


B'R 
Ci  ^  —  — : —  =  — 0.0000641  mho. 
4 

Ci  =  B  — -— —  =  0.001411  mho. 

The  voltage  at  the  sending-end  is  determined  as 
follows : — 

/i  (Cos  Oi.  —  j  sin  Ov)  =  89.928  —  /43-S55. 

X  (fti  4-  jbi)  =  20286  -f  yi7  8i9  volts. 
-t-  £ro  (Oi  +  ia,)  =  48361  4-  i  4Q27  volts. 

£„  =  68647 -t-y22  746. 

=  72319  /i8°  19'  56'  volts. 

The  current  at  the  sending-end  may  be  determined 
as  follows: — 

/l  (Cos  0L  —  J  sin  6l)  =  89.928  —  ;43  555- 

X   (0,  +  jOt)  =  +76.003  —  727.700  amperes. 
+  £..  (C,  +  /CO  =  —  3849  +  ;84.7i8  amperes. 

I,  =  72.154  -t-  ;57-oi8. 

=:  91.962  738°  19'  03"  amperes. 

The  above  results  check  exactly  with  those  pre- 
viously obtained  by  impedance  calculations.  This 
agreement  indicates  that  the  nominal  tt  solution  may,  if 
desired,  be  used  with  complex  quantities,  assuming 
values  for  the  auxiliary  constants  as  indicated  in  Table 

Q- 

Convergent  Series  Expression — Table  Q  indicates 
that  the  nominal  tt  solution  is  equivalent  to  using  the 
following  values  for  the  auxiliary  constants  in  the  con- 
vergent series  form  of  solution, — 

.^-=(.  +  t?).  ^'  =  z,   ^..-O  +  tJ) 

We  will  now  show  that  the  above  expressions  yield 
the  same  values  for  the  auxiliary  constants  as  given  in 
Table  Q.     From  chart  XI  the  following  values  corre- 
sponding to  problem  X  are  taken. 
ZY  =  —0.389187  -f  /0.164115 
therefore, 

A'  =  i.ooooooo 

— O.I94S935  +  /  0.0820575 

A'  =  0.8054065  -1-  y 0.0820575 

B'  =  105  -t-  y249 
C  =  i.oooooo 

—0.0972967  +  y  0.0410287 


=  Y  (0.9027033  -f-  y  0.0410287) 

C  =  — 0.0000641  -|-  yo.001411 
Thus  the  values  for  the  auxiliary  constants  as  de- 
termined by  the  above  incomplete  convergent  series  ex- 
pression check  with  those  as  determined  above  from  the 
equations  in  Table  Q. 


COMPARISON  OF  VARIOUS  METHODS 


115 


MIDDLE  CONDENSER  OR  NOMINAL  T  METHOD 

THIS  METHOD  assumes  that  the  total  capaci- 
tance of  the  circuit  may  be  concentrated  at  its 
middle  point.  In  such  a  case  the  entire  charging 
current  would  flow  over  half  of  the  circuit.  The  re- 
sistance and  the  reactance  on  each  side  of  the  capaci- 
tance or  condenser  is  equal  respectively  to  half  the  total 
conductor  resistance  and  conductor  reactance. 

From  an  inspection  of  the  diagram  of  such  a  cir- 
cuit, Fig.  36,  it  is  evident  that  two  calculations  will  be 
required.  Starting  with  the  known  receiving-end  con- 
ditions, the  conditions  at  the  middle  of  ihe  circuit  are 
first  calculated  by  the  simple  impedance  method.  To 
these  calculated  results  the  current  consumed  by  the 
condenser  shunted  across  the  middle  of  the  circuit  must 
be  vectorially  added.  This  will  give  the  load  condition 
at  the  middle  of  the  circuit  from  which  the  sending-end 
conditions  may  be  calculated. 

Solution  by  Impedance  Method — The  diagram  of 
ctjnnections  and  the  corresponding  graphical  vector 
solution  for  problem  X  by  the  nominal  T  method  is 
indicated  by  Fig.  56.  The  electrical  conditions  at  the 
middle  of  the  circuit  may  be  determined  as  follows: — 

/r—  =  99.92  X  52.5  =  5246  volts  (resistance  drop) 


X 


99.92  X  124.5  =  12440  volts  (reactance  drop) 


Emu       —   1/  (60046  X  0.9  +  5246)-  +  (60046  X  0.4J59  +   12440)' 

=  70  753  X33°  04'  36"  to  current  vector  OD 
—  70  753  /7°  14'  05"  to  vector  of  reference  OR 

The  current  consumed  by  the  condenser  (zero 
leakage  assumed)  leads  the  voltage  OM  at  the  middle 
of  the  circuit  by  90  degrees  and  is : — 

/c  =  0.001563  X  70753  =  110.587  amperes 

The  voltage  consumed  by  the  condenser  current 
flowing  back  to  the  sending-end  is : — 

/c — =  110.587  X     52.5  =     5806  volts  (resistance  drop) 

=  FC 


/c  —  =  110.587  X  124.5 


percent.     The  voltage  at  the  sending-end  will  therefore 
be:— 

£,„  =   i/(57j8o  X  0.77831  +  5246)'  +  (57280  X  0.62788  +  12440)' 

=  69  467  /44°  10'  14"  volts  to  vector  OD 

=  69467  y/iS"  19'  43"  volts  to  vector  of  reference  OR 

If  desired,  the  receiving-end  current  and  the  con- 
denser current  may  be  combined  and  the  corresponding 
impedance  triangle  for  the  sending-end  half  of  the  cir- 
cuit constructed  on  the  end  of  vector  OM  as  indicated 
by  the  dotted  lines. 

The  current  at  the  sending-end  may  be  determined 

as  follows: — 

OB  =  99.92  cos  33°  04'  36"  =^  83.727  amperes. 
BD  =  99.92  sin  33°  04'  36"  =  54532  amperes. 
BN  —  110.587  —  54.532  —  56.055  amperes. 

I.  =  ON  =  V  (83.727)'  +   (56.055)' 

=  100.76  /33°  48'  06"  amperes  to  vector  OB. 
■=  100.76  /'41''  02'   11"  amperes  to  vector  of 
reference  OR. 

The  current  at  the  sending-end  leads  the  voltage  at 
the  sending-end  by  the  angle  41°  02'  11"  —  18°  19'  43"' 
=  22°  42'  28",  which  corresponds  to  a  power-factor  at 
.  the  sending-end  of  92.25  percent  leading. . 

The  power  at  the  sending-end  is: — 
Kv-a„i  =  100.76  X  69467  =  7000  kv-a. 
Kw,u     =  7000      X  0.9225  =  6457  kw. 
Lossn    =6.^57     —  5400     =  1057  kw. 
Solution  by  Complex  Quantities — From   table  Q 
the  auxiliary  constants  corresponding  to  the  nominal  T 
method  of  solution  are  found  as  follows: 


13768  volts  (reactance  drop) 
=  FM 

The  voltage  vector  OC  upon  which  the  impedance 
triangle  corresponding  to  the  receiving-end  load  cur- 
rent /r  =  /l  flowing  over  the  sending-end  half  of  the 
circuit  is  constructed,  may  be  found  as  follows: — 
OC  =  V'  (70753-  i3  768)="T-l8o6= 

=  57  280  /s°  49'  03"  volts  to  vector  OM 

=  57  280  /iz°  03'  08"  volts  to  vector  of  reference  OR 

The  voltage  OC  leads  the  receiving-end  current 
OD  by  the  angle  33°  04'  36"  +  5°  49'  03"  =  38°  53'  39" 
which  angle  corresponds  to  a  power-factor  of  77-831 


O,  =  I  — 
RB 

b,  =  X 


XB_ 
2 

RXB 


2 
V 


=  0.805  406  5 
=  0.082  057  5 
=  84.5677 


— (X'  —  R')  =  229.081 

c.  =  0 

ci  ■=  B  =  0.001  563 

The  voltage  at  the  sending-end  is  obtained  as  fol- 
lows : — 

Ik  (cos  Sr  —  /  sin  Or)  =  89.928  —  / 43-554 

X  (fci  -f-  ih)  =  17582  +  yi69i8 
+Etu  (flj  +  j  a,)  =  48361  -f-  ;4927 

£.a  =  65943  +  /21845 
=  69467  /i8°  19'  43" 

The  current  at  the  sending-end  may  be  calculated 
as  follows: — 

/u  (cos  Sn  —  j  sin  »r)  ==  89.928  —  /  43-554 

X  (Oi  -f  /oj)  =  76.0026  —  /  27.6994 
+  Etu  (ci  +  jc)  =      o       +  y  93-8519 

/,  =  76.0026  4-  y  66.1525 

=  100.76  /4i°  02'  11"  amperes 


ii6 


COMPARISON  OF  VARIOUS  METHODS 


The  above  results  check  with  those  previously  ob- 
tained by  impedance  calculations.  This  agreement  in- 
dicates that  the  nominal  T  solution  may,  if  desired,  be 
made  by  complex  quantities,  assuming  values  for  the 
auxiliary  constants  as  indicated  in  Table  Q. 

Convergent  Series  Expression — Table  Q  indicates 
that  the  nominal  T  solution  is  equivalent  to  using  the 
following  values  for  the  auxiliary  constants  in  the  con- 
vergent series  form  of  solution: — 


=^('-^) 


B 

C  =  Y 

Comparing  the  above  expressions  for  the  auxiliary 
constants  with  the  complete  expression  yielding  rigor- 
ous values  the  following  difference  may  be  noted. 

For  auxiliary  constant  A'  the  first  two  terms  in  the 
complete  series  for  the  hyperbolic  cosine  are  used  and 


expression!!    check   exactly   with    those   as   determined 
above  from  the  equations  in  Table  Q. 

THREE  CONDENSER  METHOD 

This  method  (proposed  by  Dr.  Chas.  P.  Steinmetz) 
assumes  that  the  admittance  of  the  circuit  may  be 
lumped  or  concentrated  across  the  circuit  at  three 
points,  one-sixth  being  localized  at  each  end  and 
two-thirds  at  the  middle  of  the  circuit.  This  is  equiva- 
lent to  assuming  that  the  electrical  quantities  are  dis- 
tributed along  the  circuit  in  a  manner  represented  by 
the  arc  of  a  parabola.  It  is  evident  that  this  method 
more  nearly  approaches  the  actual  distribution  of  the 
impedance  and  the  admittance  of  the  circuit  than  any 
of  the  three  previously  described  localized  admittance 
methods,  and  therefore  yields  more  accurate  results. 

From  an  inspection  of  the  diagram  of  such  a  cir- 
cuit, Fig.  57,  it  will  be  evident  that  it  is  necessary  to 
calculate  the  performance  of  the  two  halves  of  the  cir- 


PROBUEM  'X" 

KV-Arn-    6,000  KVA       Ern  •=  80'<'*8  volts 

*  KWrn  -    6,400  KW  PFr  -  90*  LAQQINQ 

\  Ip -89.92  AMPERES       F  -  60  CYCLES 

SENDING    R  X  R  X,''E°E'^"*° 

^  ~  ~  2        END 


LINEAR  CONSTANTS 
R  -  106  OHMS 
X  -  249  OHMS 
B  -  0.001663  MHO 
Q  -  0  (SO  THAT  Y-B) 


NEUTRAL 
RESULTS  CALCULATED  BY 


RIGOROUS  SOLUTION 

MIDDLE 

CONDENSER 

METHOD 

%  ERROR 

EsN-  70.662  VOLTS 

69,467  VOLTS 

-  1.68% 

Is-  94.76  AMPS. 

100.76  AMPS. 

+  6.34% 

PFs- +  .93.42% 

*  92.26% 

-  1.26% 

LOSS  N- 866  KW 

1067  KW 

«■  23.63% 

Ir  9  "  12,440  VOLTS 


>  6246  VOLTS 


12,440  VOLTS 


F0-|c5    -  6806  VOLTS 
-    -■  13.768  VOLTS 


FIG.   S6 — NOMINAL  T  OR   MIDDLE  CONDENSER   METHOD 


all  terms  beyond  omitted.  For  auxiliary  constant  B' 
the  first  two  terms  of  the  complete  series  are  also  used 
except  that  the  coefficient  of  the  second  term  is  given 
as  J4.  whereas  in  the  complete  series  it  is  i/6.  Auxi- 
liary constant  C  is  equivalent  to  the  first  term  only  of 
the  complete  expression. 

We  will  now  show  that  the  above  expressions  yield 
the  same  values  for  the  auxiliary  constants  as  given  in 
Table  Q.  From  Chart  XI  the  following  values  corre- 
sponding to  problem  X  are  taken : — 

Z  =        105  +  /240 
Z  y  =  —  0.389187  -f-  ;o.i64ii5 
Therefore /4 '  =       i.oooooo 

—  O.I  945  935  +  ;" 00 820  S75 
A'  =  +  0.8054065  -I-  ;■  0.0 820 575 
B'  =        I.oooooo 

—  0.09 729 675  -I-  ;" 004 102 875 

=  z  (0.90270325  +  y  0.04 102875) 

B'  =      84.5677  4-  /  229.081 
C  =       o  +  /o.ooi  563 
Thus  the  values  for  the  auxiliary  constants  as  de- 
termined by  the  above   incomplete   convergent   series 


cuit  in  order  to  arrive  at  the  sending-end  voltage  and 
an  additional  calculation  will  be  required  to  determine 
the  sending-end  current,  power  and  power-factor. 

Solution  by  Impedance  Method — The  diagram  of 
connections  and  corresponding  graphical  vector  solu- 
tion for  problem  X  by  the  three  condenser  method  is 
mdicated  by  Fig.  57.  The  charging  current  consumed 
by  the  condenser  (zero  leakage  assumed)  at  the  re- 
ceiving-end leads  the  receiving-end  voltage  by  90  de- 
grees and  is: — 
O.OOI  563 


/cr    = 


X  60046  =  15.642  amperes. 


The  current  per  conductor  for  the  receiving-end 
half  of  the  circuit  is : — 

/r  =  1/  (99.92  X  0.9)"  -I-  (9992  X  0.4359  —  15-642)' 
=  94.16  \i7°  14'  38"  amperes 
PF.  =  Cos  \i7°  14'  38"  —  95.505  lagging 

The  voltage  consumed  by  the  resistance  and  the 
reactance  per  conductor  between  the  receiving-end  and 
the  middle  of  the  circuit  is: — 


COMPARISON  OF  VARIOUS  METHODS 


U7 


/r— =  9416  X  525  =  4943-4  Volts  (resistance  drop) 

,   X 

/r~=  94-i6  X   124.5   =  11723  Volts  (reactance  drop) 

The  voltage  at  the  middle  of  the  circuit  is  from 
(30):-        

£mi.  =  y  (60046X  0.95505  +  4943-4)'+  (60046X0.29644+  11723)" 

=  68933  /25°  21'  33"  volts  to  current  vector  OP 
=  68933  y8°  06'  55"  volts  to  vector  of  reference  OR 

The  charging  current  consumed  by  the  condenser 
(zero  leakage  assumed)   at  the  middle  of  the  circuit 
leads  the  voltage  at  the  middle  of  the  circuit  by  90  de- 
grees and  is: — 
o.ooi  563 


The  current  at  the  sending-end  of  the  circuit  may 
be  determined  as  follows: — 

OS  =  Cos  10°  19'  07"  X  90.73  =  89.2624 
VS  =  Sin  10°  19'  07"  X  90-73  =  16.2516 
A^5"  =  16.2516  +  18.3777  =  346293  amperes. 
I.=\/  89.2624'  +  34-6293' 

=  95-744  /2i°  12'  13"  to  voltage  vector  OS. 

=  95-744  /^39°  18'  56"  to  vector  of  reference  OR. 

Kv-a,n  =  95-744  X  70-548  =  6755  kv-a 
PF.  =  Cos  (39°  18'  56"  —  18°  06'  43") 

=  Cos  21°  12'  13"  =  93.23  percent  leading 
Kw,n  =  6755  X  0.9323  =  6298  kw 
LosSu  =  6298  —  54CX3  =:  898  kw 
5400  X  100 


Eff. 


6298 


85-75  percent. 


/.„ 


-X68933  =  71.828  amperes. 


Solution  by  Complex  Quantities — From  Table  Q 
the  auxiliary  constants  corresponding  to  the  three  con- 
The  current  per  conductor  for  the  sending-end  half     denser  method  of  solution  are  found  to  be:— 
of  the  circuit  may  be  determined  as  follows: —  XB        B' 


OT  =  Cos  25°  21'  33"  X  94-16  =  85.0867  amperes. 

PROBLEM 'X' 


O, 


-~  + 


36 


(X'  —  i?')  =  0.808866 


KV-Aln"  ''""^  '*^*        Ern"  60,046  VOLTS 
II  -  99.92  AMPERES   F=  80  CYCLES 


LINEAR  CONSTANTS 
R  -  106  OHMS 
X  -  248  OHMS 
B  -  0.001683  MHO 
O  -  0  (80  THAT  Y"B) 


RIGOROUS  SOLUTION 

THREE 

CONDENSER 

METHOD 

%  ERROR 

EsN-  70,662  VOLTS 

70,648  VOLTS 

-0.16% 

Is-  94.75  AMPS. 

96.744  AMPS. 

•H.06% 

PFs"  ♦  93.42% 

•1-  93.23% 

-0.21% 

LOSSn- ''''<* 

898  KWf 

♦  6.03% 

l6.e42AMP8.^ 


riG.  57 — DR.  CHAS.  P.   STEINMETZ  S  THREE  CONDENSER  METHOD 


TP  =  Sin  25°  21'  33"  X  94-i6  =  40.3278  amperes. 
TV  =  71.828  —  40.3278  ^  31.5002  amperes. 

/m  =  1      85.0867-'  -I-  31.5002= 

=  90.73  /20°  18'  55"  amperes  to  voltage  vector  OM  al 

tniddle. 
=  90.73  /28°  25'  50"  to  vector  of  reference  OR 

The  voltage  consumed  by  the  resistance  and  the  re- 
actance per  conductor  between  the  middle  and  sending- 
end  of  the  circuit  is: — 

/m—     X  90.73  X  52.5  =  4763-3  volts  (resistance  drop) 

,     X 

/m  —'    X  90.73  X  124.5  =  II  296  volts  (reactance  drop) 

The  voltage  at  the  sending-end  from  (40)  is: — 

£«n  =  V  (68933  X  0.93779  +  4763-3)'  +  (68933  X  0.34719  —  11  296)' 

=  70548  \io°  19'  07"  volts  to  current  vector  OV 

=  70548  /l8°  c6'  43"  volts  to  vector  of  reference  OR 

The  charging  current  consumed  by  the  condenser 
(zero  leakage  assumed)  at  the  sending-end  of  the  cir- 
cuit leads  the  voltage  at  the  sending-end  by  90  degrees 
and  is: — 

0.001563 
6 


b,  =  R 
b, 
c»  =  — 

c,  —  B 


RB  _  RXB'- 
2         '18 
RXB 


0.0785091 
=  91.3785 
X  -  -|   (X'  -  R')  =  235.7208 


5  RB'        RXB 
+ 


0.0000347 


36      ^     108      " 

5  XB'  B' 

^6~  +  116  <^'  -  '^'>  =  +  ooo'4794 


These  values  for  the  auxiliary  constants  are  in 
close  agreement  with  the  rigorous  values. 
/l  (Cos  0l  —  ;"  Sin  Sl)  X  (b,  +  jb,) 

—  18  484 -I- y  17  218 

Em  (a,   +  JO,)  =  48569  -I-  y  4714 


£„  =  67  053  +  j  21 932 

=  70  548  /iS"  06'  43"  volts 

The  current  at  the  sending-end  is : — 

/l  (Cos  Al  —  y  Sin  Sl)  X  (a,  +  i  Oj) 
=      76.159  —  y  28.170 

£r.  (c,  -1-  y  c)  =  —  2.084  -I-  y  88.832 


u.  = 


X  70548  =  18.3777  amperes. 


I,  =   74075  +  y  60.662 

=      95-744  /i9°  18'  56"  amperes 
By  comparing  these  results  with  those  obtained 


118 


COMPARISON  OF  VARIOUS  METHODS 
CHART  XXII— COMPARISON  OF  RESULTS  BY  VARIOUS  METHODS 


a 
0 
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LOAD    AT 

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AS  DETERMINED  BY 

CON  STAN 

RECEIVING-END 

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*It  would  be  commercially  impractical  to  transmit  such  small  amounts  of  power  some  of  the  extreme  dis- 
tances indicated  by  the  tabulation.  The  problems  are  stated  simply  for  the  purpose  of  illustrating  in  an  approxi- 
mate manner  the  effect  distance  of  transmission  has  upon  the  voltage  drop  as  calculated  by  various  methods. 


COMPARISON  OF  VARIOUS  METHODS 


119 


by  the  impedance  method  of  procedure,  it  will  be  seen 
that  they  are  in  exact  agreement. 

Convergent  Series  Expression — Dr.  F.  E.  Pemot 
in  "Electrical  Phenomena  in  Parallel  Conductors,"  Vol. 
I,  shows  that  the  above  described  three  condenser  solu- 
tion is  equivalent  to  using  the  following  values  for  the 
auxiliary  constants  in  the  convergent  series  form  of 
solution : — 

A'  =  (^1  +  -^^  + 


B" 


2 


36 


C 


36  ^  216  / 
Comparing  the  above  expressions  for  the  auxiliary 
constants  with  the  complete  expressions  yielding  rigor- 
ous values,  the  following  differences  may  be  not^d. 
For  constant  A'  the  first  two  terms  are  the  same  as  in 
the  complete  series,  but  the  third  term  is  less  than  in 
the  complete  series,  and  all  terms  beyond  the  third  are 
omitted.  For  constant  B'  the  first  two  terms  are  the 
same  as  in  the  complete  series,  but  all  terms  beyond  the 
second  are  omitted.  For  constant  C  both  the  ZY  and 
the  Z-  Y=  terms  are  smaller  than  in  the  complete  series 
and  all  terms  beyond  the  third  are  omitted. 

The  above  expressions  yield  the  same  values  for 
the  auxiliary  constants  as  given  in  Table  Q.  Thus 
from  chart  XI,  the  following  values  corresponding  to 
problem  X  are  taken: — 

ZY  =  —  0.389187  +  j  0.164115 
Z^Y"  =  +  0.124532  —  }  0.127742 
Therefore 

A'  =        i.oooooo 

—  0.194593  +  J  0.0820575 
0.003459  —  y  0.0035484 


A'   =   0.808866  -I-  y  0.0785091 
B'   =   I.oooooo 

—  0.0648645  +  j  0.0273525 


z  (0.9351355  +  y  0.0273525) 
B'  =   913785     +  y  235.7208 

C  =        I.oooooo 

—  0.0540538  +  j  0.0227938 
+  0.0005765  —  y  0.0005914 


y  (0.9465227  -f-  /  0.0222024) 
c  =  —  0.0000347  -1-  y  0.0014794 

It  will  be  seen  that  the  above  convergent  series  ex- 
pression for  the  auxiliary  constants  check  exactly  with 
those  as  determined  by  the  equations  in  Table  Q. 

COMPARATIVE   ACCURACY    OF   VARIOUS    METHODS 

In  order  to  determine  the  inherent  error  in  various 
methods  of  solution,  when  applied  to  circuits  of  in- 
creasing length;  also  for  frequencies  of  both  25  and 
60  cycles,  64  problems  were  solved.  These  problems 
embrace  thirty-two  25  cycle  circuits,  varying  in  length 
from  20  to  500  miles  and  in  voltage  from  10  000  to 
200000  volts.  Fixed  receiving-end  load  conditions 
were  assumed  for  unity,  and  also  for  80  percent  power- 
factor  lagging.  These  same  problems  were  also  solved 
for  a  frequency  of  60  cycles. 

These  64  problems  with  corresponding  linear  con- 
stants and  assumed  load  conditions  are  stated  on  Chart 


XXII.  This  is  followed  by  columns  in  which  have 
been  tabulated  the  error  in  voltage  at  the  sending-end 
of  these  circuits  as  determined  by  nine  different 
methods.  The  errors  are  expressed  in  percent  of  re- 
ceiving-end voltage.  Obviously  the  inherent  error 
corresponding  to  various  methods  will  vary  widely  for 
conductors  of  various  resistances  and  to  some  extent 
for  different  receiving-end  loads.  The  tabulated  values 
should  therefore  be  looked  upon  as  comparative  rather 
than  absolute  for  all  conditions. 

Rigorous  Solution — The  column  headed  "Rigorous 
Solution"  contains  values  for  the  sending-end  voltage 
which  are  believed  to  be  exact.  These  values  were  ob- 
tained by  calculating  values  for  the  auxiliary  constants 
by  means  of  convergent  series  and  then  calculating  the 
performance  mathematically.  The  calculations  were 
carried  out  to  include  the  sixth  place  and  terms  in  con- 
vergent series  were  used  out  to  the  point  where  they 
did  not  influence  the  results. 

The  first  values  calculated  were  checked  by  a 
second  set  of  values  calculated  independently  at  an- 
other time  and  where  differences  were  found  the  cor- 
rect values  were  determined  and  substituted.  This 
corrected  list  of  values  was  again  checked  by  a  third 
independent  calculation.  It  is  therefore  believed  that 
the  values  contained  in  this  column  are  exact,  repre- 
senting 100  percent. 

Semi-Graphical  Solution— The  next  column  con- 
tains the  error  in  the  results  as  derived  by  the  combi- 
nation of  an  exact  mathematical  solution  for  the 
auxiliary  constants  and  a  graphical  solution  from  there 
on.  This  combination  gave  results  in  which  the  maxi- 
mum error  does  not  exceed  eight  one  hundredths  of  one 
percent  of  receiving-end  voltage  for  either  frequency. 
In  other  words,  since  the  values  for  the  auxiliary  con- 
stants used  in  this  method  were  exact,  the  maximum 
error  of  eight  one  hundredths  of  one  percent  occurs  in 
the  construction  and  reading  of  the  graphical  construc- 
tions. 

Complete  Graphical  Solution — This  solution  em- 
ploys Wilkinson's  charts  for  obtaining  graphically  the 
auxiliary  constants,  the  remainder  of  the  solution  being 
also  made  graphically  as  previously  described.  It  will 
be  seen  that  the  maximum  error  as  obtained  by  this 
complete  graphical  solution  is  seven  hundredths  of  one 
percent  for  the  25  cycle  and  twenty-five  hundredths 
of  one  percent  for  the  60  cycle  circuits.  These  errors 
represent  the  combined  result  of  various  errors.  First 
there  is  a  slight  fundamental  error  in  the  basis  upon 
v.'hich  the  Wilkinson  Charts  are  constructed  when 
used  for  circuits  employing  conductors  of  various  sizes 
?nd  spacings,  the  introduction  of  this  error  making 
possible  the  simplification  attained.  Then  there  is  the  in- 
herent limitation  of  precision  obtainable  in  the  construc- 
ton  and  reading  of  the  charts  and  vector  diagrams. 

These  results  show  that  the  inherent  accuracy  of 
this  simplified,  all  graphical  solution  is  sufficiently  ac- 
curate for  all  practical  power  circuits  up  to  300  miles 
long. 


120 


COMPARISON  OP  VARIOUS  METHODS 


Dwight's  "K"  Formulas — The  high  degree  of  ac- 
curacy resulting  by  the  use  of  H.  B.  Dwight's  "K" 
formulas  should  be  noted.  This  error  is  a  maximum 
of  eleven  hundredths  of  one  percent  for  these  32 
twenty-five  cycle  problems.  The  statement  is  therefore 
justified  that  these  "K"  formulas  are  sufficiently  accur- 
ate for  all  25  cycle  power  circuits. 

For  the  60  cycle  problems  the  maximum  error  by 
the  "K"  formulas  for  problems  up  to  and  including  200 
miles  is  one-fourth  of  one  percent  of  receiving-end  volt- 
age. For  300  mile  circuits  this  error  is  one-half  of  one 
percent  and  increases  rapidly  as  the  circuit  exceeds  300 
miles  in  length.  The  accuracy  of  the  "K"  formulas  for 
60  cycle  circuits  is  therefore  well  within  that  of  the 
assumed  values  of  the  linear  constants  for  circuits  up 
to  approximately  300  miles  in  length. 

The  "K"  formulas  are  based  upon  the  hyperbolic 
formula  expressed  in  the  form  of  convergent  series. 
In  the  development  of  these  formulas,  use  was  made  of 
the  fact  that  the  capacitance  multiplied  by  the  reactance 
of  non-magnetic  transmission  conductors  is  a  constant 
quantity  to  a  fairly  close  approximation.  This  assump- 
tion has  enabled  the  "K"  formulas  to  be  expressed  in 
comparatively  simple  algebraic  form  without  the  use 
of  complex  numbers.  To  those  not  familiar  or  not  in 
position  to  make  themselves  familiar  with  the  operation 
of  complex  numbers,  such  as  is  used  in  the  convergent 


scries  or  hyperbolic  treatments,  the  availability  of  the 
Dwight  "K"  formulas  will  be  apparent.* 

Localised  Capacitance  Methods — The  next  four 
columns  contain  values  indicating  the  error  in  results 
as  determined  by  the  four  different  localized  capaci- 
tance methods  previously  described  in  detail.  It  is  in- 
teresting to  note  the  high  degree  of  accuracy  inherent 
in  Dr.  Steinmetz's  three  condenser  method.  It  is  also 
interesting  to  note  that  three  of  these  methods  over 
compensate  (that  is,  give  receiving-end  voltages  too 
low)  and  one  (the  split  condenser  method)  gives  under 
compensation. 

Impedance  Method— Th^  values  of  the  sending-end 
voltage  as  obtained  by  the  impedance  method  (which 
takes  no  account  of  capacitance)  are  always  too  high 
when  applied  to  circuits  containing  capacitance.  The 
results  by  this  method  are  included  here  simply  to  serve 
as  an  indication  of  how  great  is  the  error  for  this 
method  when  applied  to  circuits  of  various  lengths  and 
frequencies  of  25  and  60  cycles.  Some  engineers  pre- 
fer to  use  this  method  for  circuits  of  fair  length  and 
allow  for  the  error.  These  tabulations  will  give  an 
approximation  of  the  necessary  allowance  to  be  made. 

*Thesc  have  been  included  with  much  other  valuable  ma- 
terial in  "Transmission  Line  Formulas"  by  H.  B.  Dwight,  pub- 
lished by  D.  Van  Nostrand  Co.  of  New  York  City. 


CHAPTER    XIII 

CABLE  CHARACTERISTICS 

Heating  Limits  for  Cables 


THE   MAXIMUM    safe-limiting   temperatures   in 
degrees  C  at  the  surface  of  conductors  in  cables 
is  given   in   the   Standardization   Rules  of   the 
A.  I.  E.  E.  (1918)  as  follows: — 

For  impregnated  paper  insulation  (85 — E) 
For  varnished  cambric   (75 — E) 
For  rubber  insulation  (60—0.25  E) 

Where  E  represents  the  effective  operating  e.m.f. 
in  kilovolts  between  conductors  and  the  numerals 
represent  temperature  in  degrees  C.  Thus,  at  a  work- 
ing pressure  of  5  kv,  the  maximum  safe  limiting 
temperature  at  the  surface  of  the  conductors  in  a 
cable  would  be: — 

For  impregnated  paper  insulation  (80  degrees  C) 
For  varnished  cambric  insulation  (70  degrees  C) 
For  rubber  compound  insulation  (58.75  degrees  C) 

The  acttial  maximum  safe  continuous  current  load 
for  any  given  cable  is  determined  primarily  by  the  tem- 
perature of  the  surrounding  medium  and  the  rate  of 
radiation.  This  current  value  is  greater  with  direct 
than  with  alternating  current  and  decreases  with  in- 
creasing frequency,  being  less  for  a  60  cycles  than 
for  25  cycles.  The  carrymg  capacity  of  cables  will 
therefore  be  less  in  hot  climates  than  in  cooler  climates 
and  will  be  considerably  increased  during  the  winter. 

Cables  immersed  in  water,  carry  at  least  50  per- 
cent more  than  when  installed  in  a  four-duct  line,  and 
when  buried  in  the  earth  15  to  30  percent  more  than 
in  a  duct  line,  depending  upon  the  character  of  soil 
moisture,  etc.  Circulating  air  or  water  through 
conduits  containing  lead  covered  cables  will  increase 
their  capacity.  From  the  above  it  is  evident  that  no 
general  rule  relative  to  carrying  capacity  can  be  formu- 
lated to  apply  in  all  cases,  and  it  is  necessary,  there- 
fore, to  consider  carefully  the  surroundings  when  de- 
termining the  size  of  cables  to  be  used. 

The  practicability  of  tables  which  specify  car- 
rying capacity  for  cables  installed  in  ducts  will  gen- 
erally be  questioned,  for  the  reason  that  operating  con- 
ditions are  frequently  more  severe  than  those  upon 
which  table  values  are  based.  A  duct  line  may  op- 
erate at  a  safe  temperature  throughout  its  entire  length, 
except  at  one  isolated  point  adjacent  to  a  steam  pipe 
or  excessive  local  temperatures  due  to  some  other 
cause.  If  larger  cables  are  not  employed  at  this  point, 
burnouts  may  occur  here  when  the  remainder  of  the 
cable  line  is  operating  well  within  the  limits  of  safe 
operating  temperature.  The  danger  in  using  table 
values  for  carrying  capacity  without  carefully  consid- 
ering the  condition  of  earth  temperatures  throughout 
the  entire  duct  length  is  thus  evident. 


HEATING  OF  CABLES TABLE  XXIV 

The  basis  upon  which  the  data  in  Table  XXIV 
has  been  calculated  is  covered  by  foot  notes  below  the 
table.  The  kv-a  values  are  determined  from  the  cur- 
rent in  amperes  and  are  based  upon  30  degree  C  rise 
and  a  maximum  of  3CXX)  volts.*  Expressing  the  car- 
rying capacity  of  cables  in  terms  of  kv-a  (corrected 
for  the  varying  thickness  of  insulation  required  for 
various  voltages)  may  be  found  more  convenient 
than  the  usual  manner  of  expressing  it  in  amperes. 
It  will  be  noted  that  the  kv-a  values  of  the  table  are 
on  the  basis  of  a  four-duct  line  and  that  for  more 
than  four  ducts  in  the  line  the  table  kv-a  values  will 
be  reduced  to  the  following : — 

For  a  4  duct  line — 100  percent. 
For  a  6  duct  line —  88  percent. 
For  an  8  duct  line —  79  percent. 
For  a  ID  duct  line —  71  percent. 
For  a  12  duct  line —  63  percent. 
For  a  16  duct  line —  60  percent. 

When  applied  to  all  sizes  of  cables,  the  above 
values  are  only  approximate.  The  reduction  of  car- 
rying capacity  caused  by  the  presence  of  many  cables 
is  more  for  large  cables  than  for  small  ones.  Also, 
where  load  factors  are  small,  the  reduction  due  to  the 
presence  of  many  cables  is  less  than  the  value  assigned, 
although  the  carrying  capacity  of  a  small  number  of 
cables  is  only  slightly  affected. 

REACTANCE  OF  THREE-CONDUCTOR  CABLES 

Tables  XXV  and  XXVI  contain  values  for  the 
inductance,  reactance  and  impedance  of  round  three- 
conductor  cables  of  various  sizes  and  for  the  thick- 
nesses of  insulation  indicated.All  values  in  the  tables 
are  on  the  basis  of  one  conductor  of  the  cable  one  mile 
long. 

The  table  values  were  calculated  from  the  funda- 
mental equation  (4), 

D 

L  =  0.08047  -f-  0.741  logu—S- 

where  L  =  the  inductance  in  millihenries  per 
mile  of  each  conductor,  R  the  actual  radius  of  the 
conductor  and  D  the  distance  between  conductor  cen- 
ters expressed  in  the  same  units  as  R.  As  indicated  in 
Section  I,  under  Inductance,**  this  formula  has  been 
derived  on  the  basis  of  solid  conductors.  In  the  case 
of  cables,  the  effective  radius  is  actually  slightly  less 
than  that  of  the  stranded  conductor.     The  values  for 


♦These  current  values  are  taken  from  General  Electric 
Bulletin  No.  .49302  dated  March  1917.  They  are  in  general 
slightly  higher  than  those  published  by  the  Standard  Undei^ 
ground  Cable  Company  in  their  Hand  Book  dated  1906. 

♦♦Chapter  I. 


122 


CABLE  CHARACTERISTICS 


TABLE  XXIV— CARRYING  CAPACITY  OF  INSULATED  COPPER  CONDUCTORS 

The  following  values  for  carrying  capacity  must  not  be  assumed  unless  it  is  positively  known  that  the  conditions  upon 
which  they  are  based  will  not  be  exceeded  in  service. 


THREE 

CON  DUCTO  R 

CABLES 

B  &  S  NO. 

AREA 

IN 

CIRCULAR 

MILS 

XX 

0*RH»iNG 

CftPAC'TV   IN 
-AMPERES 

Direct    - 

CURRENT 
BASED    UPON 
30«  C    «  '  S  E 

AND    A 

MAXIMUM  OF 

3000  rfOLT& 

PAPER  IN. 

8ULATION 

K.V  A     W 
LEAD  OOVERE 
THAT  ALL  DUC 

FOR  A  6 
TO  79  PER  CE 
(4  WIDE  AND  t 

HIGH    MAY  BE    TRANSMITTED   AT   THREE    PHASE   AND   THE    FOLLOWING  VOLTAGES  OVER  PAPER  INSULATED 
D  CABLES  INSTALLED  IN  A  FOUR  DUCT  LINE  WITH  30°  C  RISE  IN  TEMPERATURE  BASED  UPON  THE  ASSUMPTION 
DTS  CARRY  LOADED  CABLES  AND  UPON  A  NORMAL  EARTH  TEMPERATURE  OF  20°  C 
DUCT  LINE  THESE  K.V.A  VALUES  WOULD  BE  REDUCED  TO  APPROXIMATELY  88  PER  CENT    FOR  AN  8  DUCT  LINE 

NT    FOR  A  10  DUCT  LINE  TO  71  PER  CENT    FOR  A  12  DUCT  LINE  TO  63  PER  CENT  AND  FOR  A  I6  DUCT  LINE 
HIGH)  TO  60  PER  CENT  OF  THE  TABLE  VALUES      X  X  X  X . 

220 
VOLTS 

440 
VOLTS 

550 
VOLTS 

1100 

VOLTS 

2200 
VOLTS 

3300 
VOLTS 

4000 
VOLTS 

6000 
VOLTS 

6600 
VOLTS 

10000 
VOLTS 

II 000 
VOLTS 

12000 
VOLTS 

13200 
VOLTS 

15000 
VOLTS 

20000 

VOLTS 

22000 
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25000 
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AMPERES   DIRECT 

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CURRENT 

PAPER  INSULATED  LEAD  COVERED  CABLES  INSTALLED  IN  A  FOUR  DUCT  LINE  with  30"  C  RISE  IN  TEMPERATURE 

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FOR  A  6  DUCT  LINE  THESE  K.V.A.  VALUES  WOULD  BE  REDUCED  TO  APPROXIMATELY  88  PER  CENT    FOR 
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X  For  purposes  of  comparison  these  values  are  given  for  interior  conductors. 

XX  For  four  conductor  cables  these  ampere  values  would  be  reduced  by  12.5  percent. 

XXX  For  solid  conductors  these  ampere  ratings  would  be  reduced  by  seven  percent.  For  two  conductor  cables  made  up 
cither  round  or  flat,  they  would  be  reduced  by  15  percent.  For  two  conductor  concentric  cables  they  would  be  reduced  by  25 
percent.  They  will  also  be  reduced  in  the  case  of  the  larger  conductors  when  used  on  alternating-current  circuits  on  account 
of  skin  effect,  unless  special  cables  having  non-conducting  cores  are  used.  These  special  cables  should  be  used  for  700000 
circ.  mils  and  larger  for  60  cycle  and  i  000  000  circ.  mils  and  larger  for  25  cycle  service. 

XX  XX  For  the  higher  voltage  cables  the  kv-a  values  of  the  table  have  been  reduced  by  one  percent  for  each  2000 
volts  that  the  working  pressure  exceeds  3000  volts,  that  is  by  11  percent  for  a  25000  volt  cable.  For  insulated  aluminum 
conductors  the  safe  carrying  capacity  (based  upon  61  percent  conductivity)  is  79.3  percent  of  the  above  table  values  with  the 
same  kind  of  insulation.    These  kv-a  values  are  based  upon  the  current  in  columns  headed  by  XX  and  XXX. 


CABLE  CHARACTERISTICS 


123 


TABLE  XXV— INDUCTANCE,  REACTANCE  AND  IMPEDANCE,   AT  25  CYCLES,  PER  MILE  OF 
SINGLE  CONDUCTOR  FOR  THREE  CONDUCTOR  CABLES 


AREA 

IN 

CIRCULAR 

MILS 

B  &  S  NO. 

a:  CO 

UJ  UJ 

1-  I 

UJ  0 

S  z 

<;~ 
OS 

RESISTANCE 

PER  MILE 
IN  OHMS  )f 

INSULATION  THICKNESS  IN  64THS  OF  AN  INCH    *  • 

^   BY   ^ 

4     nv     4 
67    BY    6^ 

A    BY    ^ 

^    BY    ^ 

IND. 
M.H. 

REAC. 
OHMS 

IMP. 
OHMS 

IND. 
M.H. 

REAC. 
OHMS 

IMP. 
OHMS 

IND. 
M.H. 

REAC. 
OHMS 

IMP. 
OHMS 

IND. 
M.H. 

REAC. 
OHMS 

IMP. 
OHMS 

•Soo  ooo 
-4S0  ooo 
•^oo  ooo 

.772 
•  7XS 

.1/6 
./3.<f 
./4S 

.33S 
.340 
.343 

.O.5'30 

.0534 
.0537 

.128 
.140 
.ISS 

.349 
.3S  1 
.3S4 

.0.547 
.ossz 

.OSS4 

.129 
.140 
.iss 

.360 
.362 
.367 

.0S6S 
•  0S68 
,0576 

./27 
.14  1 
•  ISS 

.370 
.3  73 
.377 

.0.5  80 
.0J85 
.0.572 

.130 
.142 
.IS7 

3S0 ooo 
3oo  ooo 
ZSo  ooo 

.hei 
.630 
•S75 

.ihi, 

.233 

.3-J6 
.34'! 
.3S3 

.0S43. 
.0S47 
.OSS4 

.ns 
.204 

.240 

.3S7 
.361 
.366 

.0.560 

.056  7 

.0.575 

.176 
.Z04 

.240 

.370 
.3  74 
.38/ 

•  OSSI 
.0587 
.0577 

•  176 

•20s 
.240 

.3SO 
,386 
.394 

.OS76 
,060S 
.OAI9 

•  177 
.207 
.242 

oooo 
ooo 
oo 

..528 
.■^70 
.418 

.27-5' 

.357 
.362 
.369 

,osio 

.OSi7 

.OS  79 

.28/ 
.3.52 

.44/ 

.372. 
.379 
.388 

.058.5 
.OS9S 
.0607 

.Z8I 
.3.52 

.442 

.3  8  7 
.377 
.406 

.0607 
.0623 
.o637 

.282 
.352 
.44  2 

.403 
.41  1 
.42  3 

.0633 
.0645 
.0665 

.282 
•  353 
.442 

o 
/ 

.373 

.332 
.292 

.sso 

.693 
.87? 

.377 
.384 
.393 

,os9Z 
.0603 
.06/ 7 

.SS2 
.697 
.882 

.398 
.40s 

.417 

.QtXS 
,063S 
.0655 

.SS4 
.698 
.882 

.4/7 
.42? 
•441 

,o653 
.0673 
.0691 

.55-4 
.678 
.882 

.432 
.447 
.463 

.06-'7 
•  0700 
.0727 

•SS4 
•  699 
.882 

3 

.a.to 

.232 

./8-f 

li^ 

.403 
.413 
.437 

.0633 
.06  +  8 
.068-S- 

l.ll 
1.40 
2.21 

.43/ 
.442 
.470 

.06  7.5 
.069S 
.0737 

l.ll 
J.40 
2.2/ 

.454 
.4  67 
..50  / 

.07/2 
.0736 
,07SS 

/.ll 
1.4  0 
2.2/ 

.476 
.494 
.S29 

.0746 
.077S 
.0830 

I.I  1 
1.40 
2,21 

T     aw      1 
64    BY    64 

^   BY^ 

^    BY    ^ 

^byI| 

IND. 
M.H. 

REAC. 
OHMS 

IMP. 
OHMS 

IND. 
M.H. 

REAC. 
OHMS 

IMP. 
OHMS 

IND. 
M.H. 

REAC. 
OHMS 

IMP. 

OHMS 

IND. 
M.H. 

REAC. 
OHMS 

IMP. 
OHMS 

■SOO  ooo 
■4SO  ooo 
-400  ooo 

.91-* 
.772 
.728 

.116 
.129 
.I4S 

.379 

.354 
.389 

.OS9S 
.0602 
,o6lo 

.130 
.143 
.ISS 

•  389 

•  393 

•  396 

.O&IO 

.06/7 
,0622 

.  131 
.143 
.IS8 

.378 
.403 
.409 

.otss 
.0634 
.0642 

./32 

.144 
.IS9 

.407 
.4/' 
.-4'7 

.0640 
.0645 

.0  655 

•/33 
./45 
./60 

3SO  ooo 
300  ooo 
3  so ooo 

.630 
.S7S 

.166 
.l'>4 
.233 

.39s 
,399 

.409 

,06X0 
,Oi>2i, 
,0642 

.177 
.24i 

•  402 
.409 

.419 

.06  30 

.0642 

.0  6  5S 

.178 
.2oS 
.242 

.4IS 
.42  1 
.430 

.0652 
.0660 
.067S 

.178 
.2  OS 
.242 

.423 
.442 

.0664 
.0675- 
.0673 

•  179, 
.206 
.24.3 

oooo 

ooo 

oo 

•470 

.27.5 
.346 
.437 

.4  IS 
.429 
.439 

.06S2 
.0673 
.0690 

.283 
.3.5  3 
.44  3 

.427 

•  440 

•  4SS 

.0673 
.0670 
.0714 

.2  8  3 
.3  5^3 
.443 

.44  1 
.4SS 
.469 

.0690 
.0714 
.073S 

.2  84 
.354 

.443 

,452 
.466 
.483 

.0708 
.0730 
.0758 

.28S 
.3SS 

.444 

o 

i 

.373 
.332 
.292 

.SSO 
.69S 
.e79 

.4S3 
.46^ 
.483 

.0712 
.0732 
.07SS 

.SS4 

.466 
.483 

.SO  2 

.0731 
.07S7 
.0787 

.SS4 
.677 
.882 

.4SS 
.SOI 
.S2I 

.O760 
.078S 
,OS/6 

.sss 

■699 
.833 

.478 
..5/6 
.S37 

.0780 
.08/0 
.0843 

.ssi 

.700 
.883 

1 

.260 
.232 

./  84 

1,1 1 
/,40 
2.ZI 

.499 
.SIS 
.J -5  7 

.orez 
,oe/3 

.0873 

l.ll 

/,4o 

3,2  1 

.SI9 
.S3e 
•S80 

.O814 

.oe4S 

.0910 

I.I  1 
1.40 
2.ZI 

.S3  8 
.SS8 

.601 

.0  84  5 
.087.5 
.0743 

l.ll 
1.40 
2.2/ 

.SS8 

.0  8  74- 
.090s 
.077.5 

l.ll 
1.40 
2.2/ 

1     nv     II 
64    BY    64 

^BY^ 

^BYi| 

^BYli 

IND. 
M.H. 

REAC. 
OHMS 

IMP, 

OHMS 

IND. 
M.H. 

REAC. 
OHMS 

IMP. 
OHMS 

IND. 
M.H. 

REAC. 
OHMS 

IMP. 
OHMS 

IND. 
M.H. 

REAC. 
OHMS 

IMP. 
OHMS 

■Soo  ooo 
^So ooo 
^oo  ooo 

.SI4 
.772 
.72S 

.116 
.IZ9 
.MS 

.417 

.423 
.429 

.O&SS 

.066S 
.0673 

./33 
./4S 
.160 

.427 
.43/ 
.436 

.0670 

.067S 
.06S3 

./33 
•  I4S 
.I60 

.434 
.437 
.446 

.068/ 
,0670 
.0700 

./34 
./46 
.161 

.44/ 
.447 
.457 

.067/ 

.0705 

.07/7 

.I3S 
.147 
.162 

3^o ooo 
300 000 
ZSO  000 

.6S/ 
.630 
.S7S 

.166 
•  194 
.233 

.436 
.444 
.4S4 

.o6es 

.0697 

.0712 

.ISO 
.206 
.244 

.446 
.4.S6 
.4  6S 

.0700 
.07/5 
.0730 

•  ISO 

.206 

.244 

.453 

.46/ 
.47J- 

•07I0 
.0722 
.074S 

.ISO 
.207 

.24.r 

.464 
.473 
,486 

.0727 
.0742 

,0762 

■181 
.208 
.34S 

0000 
000 

00 

.SZ8 

.470 
^/■8 

•2  7.5- 
.346 
.437 

.46S 
.491 
.478 

,0730 

.QTSS 
.0780 

.28.i- 
.3SS 
.44S 

.476 
.473 
•  SIO 

.074S 
.077S 

.oeoe 

<2  8J 
.3SS 
•44S 

.486 
•  503 

,S2  1 

.0760 
.0790 
.08 16 

.2  86 
.3SS 
.44S 

.498 
.S16 
,S3S 

.0782 

.08/0 

.O840 

.287 
.356 
.446 

0 

k 

.373 
.331 
.292 

.sso 

.69S 
.879 

.SI4 
.S3I 
.SS4 

.0  90S 
.0830 
.0970 

.SS6 
.700 
.882 

.S28 
•  S4  6 
,S70 

.0838 

.OBSS 
.OS9S 

.SS6 
•  70  0 
.882 

.S39 
.SS9 
.S83 

.084S 
.0877 
.091S 

.SS6 
.700 
.883 

.SS4 
.573 
.S98 

.08  70 
.0700 

.0738 

.557 
■78%% 

3 

1 

.Zio 
.232 

./84 

/.40 
X.ZI 

.S74 
.S96 
.643 

.0900 
.093S 
,1010 

l.ll 
1.40 
2.2/ 

.S9I 
.6/3 
.til 

.0927 

•0962 

./037 

/.ll 
/.40 
2.2/ 

.606 
.627 
.678 

.09  SO 
.0783 
./<J63 

I.I  1 
/.40 
.2.2  1 

.6/8 

.0970 
.1  010 
.1090 

2.21 

1 

^BYil 

-Ifi-    RV     18 
64    BY    63 

20    BY    20 
64    ^^    64 

21   'aw    22 
f4    BY    64 

IND. 
M.H. 

REAC. 
OHMS 

IMP. 

OHMS 

IND. 
M.H. 

REAC. 
OHMS 

IMP. 
OHMS 

IND. 
M.H. 

REAC. 
OHMS 

IMP. 
OHMS 

IND. 
M.H. 

REAC. 
OHMS 

IMP. 
OHMS 

•Soo  000 
'4SO  000 
-400  000 

.914 
.773. 
.728 

.Ilk 
./Z9 
.  I4S 

.4S7 
.462 
.47' 

.07/7 
.07^S 
.0738 

./36 
./48 
./63 

.4V4 
.48/ 
.487 

•0744 
.07  S4 

•  0764 

.138 
.ISO 
.1  64 

.487" 

.476 

.SOS 

.0764 
.0778 

.0772 

•  /40 
.IS' 
.I6S 

.so  1 
.S09 
.S19 

.0785 

.0800 
,08/5 

.141 
.ISZ 
.166 

3  so  000 

300  000 
ZSO  000 

.(,81 
.630 
.S7S 

.166 
.194 
.233 

.4  SO 
'491 
.SOS 

.07S3 
.0770 
.079a. 

.182 
.XOS 
.24  6 

-4  7  6 
.SI  1 
.S24 

•  0778 
.Ot02 
,09%2 

.1  83 

.SI  3 
.S26 

.S4I 

.08O5 
.OS25 
.0  848 

.I8S 

.Zl  1 
.248 

.527 
.S4I 
.SS7 

.0830 
.0  848 

.0875 

•  IS6 
.2/2 
.247 

0000 
000 

00 

.SZ8 
.470 

.418 

.27.5- 
.3->6 
.437 

.sn^ 
,J36 
..162 

*OBIO 
.O840 
.OSiS 

.287 
.3  SI 
.446 

•S3  6 
.SSi 
•S7S 

.0840 
.08  70 
.0707 

.288 

.3.57 
.446 

.ssi 

.S7S 
.S99 

.0870 
.09OS 
.0940 

.287 
.3S8 

.447 

.573 
.572 
.618 

-0700 
.0730 
.0770 

.270 
.360 
.448 

0 
/ 
z 

.373 
.332 
.292 

.SSO 

.69S 
.■879 

.S7S 
.S98 
.623 

.0902 
.093S 
.097S 

.sss 
.700 

.8  84 

.60  / 
.623 

.647 

.o74i 

.O980 

•  ion 

.558 
.700 
.8  84 

.62/ 
.64S 
.674 

.0772 
.1010 
.1060 

.SSS 
.70/ 

•  kes 

■.itL 

.1  OOS 
.I04S 
,I08S 

.55-7 
.702 
.886 

3 

1 

.260 
.232 

./S4 

I.I  1 
J.40 
2.2/ 

.&49 
.673 

.10  19 
.loss 

.II3S 

1.41 
2,2  2 

.674 
.701 
.  7.S4 

.1060 
.1100 

.1 1  90 

I.I  1 

r.4i 

2.22 

.678 
.72  5- 
.780 

.I09S 
.1 138 

,I2XS 

1.1  1 
1.41 

2.2  Z 

'73.  1 

.1130 

•  1  no 
,1270 

1.12 
1.41 
2.2  2 

♦Resistance  based  upon  loo  percent  conductivity  at  25  degrees  C  (^77  degrees  F),  including  two  percent  allowance  for 
spiral  of  strands  and  two  percent  allowance  for  spiral  of  conductors.  For  a  temperature  of  65  degrees  C  (149  degrees  F) 
these  resistance  values  would  be  increased  15  percent. 

**The  inductance  is  in  millihenries;  the  reactance  and  the  impedance  are  in  ohms. 

The  table  values  were  derived  from  the  equation  L  =  0.08047  +  0.741  Logi„  -^  where  R  is  the  radius  of  conductor,  D 

the  distance  between  centers  of  conductors  expressed  in  the  same  terms  as  /?.  and  Z.  the  inductance  in  millihenries  per  mila 
of  each  conductor.    All  values  in  the  table  are  single-phase  and  based  upon  a  single  conductor  one  mile  long. 


124 


CABLE  CHARACTERISTICS 


TABLE  XXVI— INDUCTANCE,  REACTANCE  AND  IMPEDANCE,  AT  60  CYCLES,  PER  MILE  OF 
SINGLE  CONDUCTOR  FOR  THREE  CONDUCTOR  CABLES 


AREA 

IN 

CIRCULAR 

MILS 
B  &  S  NO. 

oc  CO 
liJ  UJ 

f-x 

UJO 

oz 

RESISTANCE 

PER  MILE 
IN  OHMS  ^ 

INSULATION  THICKNESS  IN  64THS  OF  AN  INCH     ••                                1 

^   BY-^ 

4       Qw       4 

64    BY    67 

^    BY   ^ 

^BY^          1 

IND. 
M.H. 

REAC. 
OHMS 

IMP. 
OHMS 

IND. 
M.H. 

REAC. 
OHMS 

IMP. 
OHMS 

IND. 
M.H. 

REAC. 

OHMS 

IMP. 
OHMS 

IND. 
M.H. 

REAC. 
OHMS 

IMP. 
OHMS 

Soo  ooo 
•^So  ooo 
-f  oo  ooo 

.8)4 
.772 
.728 

•  iz-r 

.145 

.338 
.340 
.3-»3 

.127 
.128 
.129 

.172 
.181 
.I9S 

.349 

.3SI 
,354 

.  13  1 
.132 
.134 

.ns 
./84 
.197 

.340 
.342 
.347 

,/34 
.137 
.138 

.178 
.199 
.201 

.370 
,373 
.377 

./40 
.141 
.142 

./82 
•J2V4 

3S0  ooo 
300  ooo 
2.SO  ooo 

.68/ 
.i30 
.S7J 

./94 
.233 

,3-,^ 
.349 
.3^3 

.130 
.132 
.13  3 

.21 1 
.235 
.2  48 

.357 
.3  4/ 
.344 

.I3S 
./3i 
.138 

.214 
.2  37 
.27/ 

.370 
.3  74 
.3  8/ 

,/40 
.141 
.144 

.2/7 
.240 
.274 

.380 
-384 
.3  94 

.143 
.14  S 
.149 

.22  0 
.244 
,277 

OOOO 

ooo 
oo 

.S28 

.470 

.418 

.275^ 
.34i, 
.437 

.3.57 
.342 
.3k9 

.I3S 
.13b 
.139 

.308 

.373 
.440 

.3  72 
.379 
,3  88 

,140 
.143 
.146 

.30f 
.375- 
.44/ 

,387 
,397 
,404 

.146 
./SO 
./53 

.3  13 
.378 
.4  44 

.403 
.4  1  1 
.423 

.152 
.ISS 
.HO 

.3/6 
.381 
.466 

o 

/ 
z 

.373 
.331 

.a?2 

.SSO 
.69S 

.879 

.377 
.384 
.393 

.142 
.I4S 
.148 

.S4>9 
.711 
.893 

.398 
.405" 
.4/7 

.ISO 
■.'1V7 

,57/ 

•  4/7 
,429 
,44/ 

./57 
./42 
.166 

.572 
.7/5 
.896 

,432 
,447 
,4  43 

.163 
./48 
./74 

.573 
.7/4 
.896 

3 

.2«o 
.232 
./8-f 

i.2i 

.403 
.4/3 
.437 

./S2 
.IS6 
.Its 

1.12 
1.41 
2.2  2 

.43/ 
.442 
.470 

./42 
.167 
.177 

I./3. 
/.4I 
2.22 

,454 
.44^ 
.SOI 

.171 
.177 
.189 

1.12 
/.41 
2.2Z 

.474 
.4  94 
,52  9 

.18a 
.186 
.200 

J.I2 
/■t9l 
2,2  2 

67   BY   -^ 

^   BY   ^ 

A    BY    ^ 

^BYi^, 

IND. 
M.H. 

REAC. 
OHMS 

IMP. 
OHMS 

IND. 
M.H. 

REAC. 
OHMS 

IMP. 
OHMS 

IND. 
M.H. 

REAC. 
OHMS 

IMP. 
OHMS 

IND. 
M.H. 

REAC. 
OHMS 

IMP 
OHMS 

•SOO  ooo 
ASo ooo 
•400  ooo 

.ZI4 
.772 
.738 

.Hi, 

.129 

.I4S 

•379 
.384 
.3  89 

,/43 

.I4S 
.147 

./84 
.194 
.20h 

.38-9 
,393 
.394 

,/4<i 
.148 
.149 

.186 
.I9S 
,208 

.398 
.403 

.409 

./50 
./52 
,/54 

•190 
.200 
.2/2 

,407 
.411 
.4/7 

,/53 
.ISS 
.IS7 

.192 
.202 
,2  30 

3^o  ooo 
300  ooo 
2S0  ooo 

.tei 

.i,30 
.S7S 

./a 

.194 
.2*33 

.395 
.399 
.409 

.149 

.ISO 

.IS4 

.222 
.24S 
.279 

.402 
.409 
.4/9 

.ISI 

•.'/st 

,224 
.24  b 
.282. 

.4/5 
,42/ 
,430 

.1S7 
,/S8 
.162 

.229 

f2i's 

.423 

•43/ 
.442 

./60 
.162 
.166 

,23/ 
,2  54 
.2  84 

OOOO 

ooo 
oo 

.470 

.4  1  8 

.37S 
.34(, 
.437 

.41S 
•421 
.439 

.IS7 
.li,2 
.Hi 

.3/8 
.383 

.447 

.427 
.440 
.4SS 

.171 

.320 
,385 
.4  7/ 

.44  1 
.4  55 
.449 

.166 
.172 
.177 

.32  3 
.388 
.473 

.452 
,444 
.483 

.170 
.176 
.182 

.32  3 
■319 

.4  74 

O 

I 

s. 

.373 
.332 
.3J2 

.SSO 
.69S 
.879 

.453 
.446 
.4  83 

.171 
.I7i 
.182 

.578 
.7/8 
.900 

.444 
.4  83 
.S02 

.nt 

.182 
.189 

,578 
.497 
.90a 

.485 
.SOI 
.521 

./83 
./89 
.19  6 

.580 
.720 
.902 

.499 
.516 
.537 

.18  9 
.195 
.202 

.582 

.72/ 
■  902 

3 

.ZbO 
.232 
.1  84 

l.ll 
1.40 
S.2I 

.499 
.SIS 
.IS  7 

./8» 
.I9S 
.210 

1.13 
/.4I 
2.2  2 

.SI  9 
.538 
.580 

.I9S 
.203 
.2/9 

/.1 3 
1.4  1 
2.22 

.538 

.SS8 

.&0  1 

.203 
.2  10 
.224 

1.13 
J. 42 
2.2  2 

.558 
.577 
.422 

.2/  / 
.218 
.234 

/./3 
/.42 
2.2  2 

e^BYi^ 

^BYi| 

^BYil 

14     nv     14 
64    BY    24 

IND. 
M.H. 

REAC. 
OHMS 

IMP. 
OHMS 

IND. 
M.H.I 

REAC. 
OHMS 

IMP.  . 
OHMS 

IND. 
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♦Resistance  based  upon  100  percent  conductivity  at  25  degrees  C  (77  degrees  F),  including  two  percent  allowance  ^r 
spiral  of  strands  and  two  percent  allowance  for  spiral  of  conductors.  For  a  temperature  of  65  degrees  C  (149  degrees  F) 
these  resistance  values  would  be  increased  15  percent. 

**The  inductance  is  in  millihenries;  the  reactance  and  the  impedance  are  in  ohms. 

The  table  values  were  derived  from  the  equation  L  =  0.08047  +  0.741  Logw-JJ  where  R  is  the  radius  of  conductor,  D 
the  distance  between  centers  of  conductors  expressed  in  the  same  terms  as  R,  and  L  the  inductance  in. millihenries  per  mile 
of  each  conductor.     All  values  in  the  table  are  single-phase  and  based  upon  a  single  conductor  one  mile  long. 


inductance,  as  determined  by  the  fundamental  formu- 
la, would  thus  tend  to  give  values  several  percent 
less  than  the  actual  when  applied  to  three-conductor 
cable  calculations.  On  the  other  hand  spiraling  the 
conductors  of  three  conductor  cables  tends  to  increase 
their  reactance  bv  several  percent.  It  may,  therefore,  be 


assumed  that  the  use  of  the  fundamental  formula  in 
the  case  of  three-conductor  cables  give  results  ap- 
proximately correct.  Skin  effect  on  the  larger  cables 
will,  however,  tend  to  decrease  the  reactance  slightly, 
particularly  at  60  cycles. 


C.IIU.E  CHARACTERISTICS 


125 


I    '  CAPACITANCE  OF  3   CONDUCTOR  CABLES 

Formulas  for  determining  the  approximate  capaci- 
tance of  three-conductor  cables  are  cumbersome.  They 
give  reasonably  accurate  results  only  in  the  case  of 
a  homogeneous  dielectric  and  in  cases  where  the  con- 
ductors are  small  compared  to  the  radius  of  the  sheath. 
They  give  inaccurate  results  in  cases  of  large  conduc- 
tors  closely   spaced.     Fig.   58*   illustrates  the  various 


MODEL 


RECIPROCAL  MODEL 


FIG.    58 — REPRESENTATION    OF    CAPACITANCES    OF    A    SYMMETRICAL 
THREE-PHASE  CABLE 

capacitances  of  a  three-conductor  cable.  Formulas 
taken  from  Russel's  "Alternating  Currents"  have  been 
combined  and  converted  to  common  logarithms  and 
are  given  below.  They  were  derived  by  the  method 
of  images  and  on  the  assumption  that  the  conductors 
are  round  and  symmetrically  spaced  with  respect  to 
the  axis  of  the  sheath. 

I 

li'  —  d'  + 


C. 


13.82  logi,   g  ^1  ^  . 


^       ,        {I73d_  ^         R'-d-  \ 

&.91  log,,  \^   ^         X  (^R' ^  K' d' +  d'YA ) 

C\..=  —^ 

R'  —  d' 
13.82  log,,  .R'd'r 


X  0.179  X  K.    (70) 


1.73'^   ., 


R' 


(R'  +  R'  (P  +  d 


"W 


X  0.179  X  K      (71) 


13-82  log,o  (  — — 

Where  — 

R      =  inside  radius  of  sheath  in  centimeters  (Fig.  59). 

r      =  radius  of  conductor  in  centimeters. 

d  =  distance  between  axis  of  conductor  and  axis  of 
sheath  in  centimeters. 

K  =:  the  dielectric  constant.  For  impregnated  paper  in- 
sulation it  varies  between  3  and  4;  for  varnished 
cambric  insulation  it  varies  between  4  and  6;  for 
rubber  insulation  it  varies  between  4  and  9. 

Ci  =:  capacitance  in  microfarads  per  mile  between  one 
conductor  and  the  other  two  conductors  plus  the 
sheath. 

Ci-,  =  mutual  capacitance  in  microfarads  per  mile  between 
any  two  conductors.  The  capacitance  to  neutral  is 
twice  this  value. 

C]2  is  used  in  determining  the  capacitance  for  various 
combinations  or  arrangements  as  explained  below. 

CAPACITANCE  AND  SUSCEPTANCE — TABLE   XXVII 

Table  XXVII  contains  values  for  capacitance  and 
susceptance  of  three  conductor  paper  insulated  cable 
for  the  various  sizes  of  conductors  and  thicknesses  of 
insulation  indicated.  All  values  are  based  upon  a  value 
for  K  of  J.5  and,  as  indicated,  a  thickness  of  insulation 
for  the  jacket  the  same  as  that  surrounding  each  con- 

*Reproduccd    from   Alexander  Russel's  "Alternating  Cur- 
rents." 


ductor.     The  values  were  calculated  by  equations  (70) 
and  (71). 

The  susceptance  values  given  for  25  and  60  cycles 
are  to  neutral.  In  calculating  the  voltage  regulation 
of  circuits,  it  is  general  practice  to  calculate  the  regu- 
lation on  the  basis  of  one  conductor  to  neutral.  The 
susceptance  between  two  of  the  conductors  would  be 
half  the  table  values  to  neutral.  The  values  for  suscep- 
tance were  calculated  from  the  equation, — 

Susceptance  to  neutral  in  micromhos  =  2  ir  f  C 
Thus  No.  o  three-conductor  cable  with  7/64  and 
7/64  insulation  has  a  capacitance  between  conductors  of 
0.195  microfarads  (0.39  microfarads  to  neutral).  The 
susceptance  to  neutral  at  60  cycles  therefore  is, — 
2  Tr  60  X  0-39  =  147  microfarads,  as  indicated  by  the 
table. 

INTER-RELATION   OF   CAPACITANCE  OF  THREE- 
CONDUCTOR  CABLES 

The  following  equations  for  determining  the  effec- 
tive capacitance  for  various  arrangements  of  the  three 
conductors  and  the  sheath  are  given  in  Russell's  "Alter- 
nating Currents." 

Capacitance  between  I  and  .?  =  J/^  (G  —  Co)  (72) 

Capacitance  between  I  and  2,  3=.  %   (Ci  —  Cu)   ....    (73) 
Capacitance  between  I  and  S  {3  and  3  insulated)  = 
(C,  —  U)    (C,  +  zC») 

(74) 


Capacitance  between  i  and  S,  2  (3  insulated)  ■= 
(C\  —  C«)    (C.  4-  C„) 


G 


Capacitance  between  I  and  S,  2,  3  ^  C, 

Capacitance  between  S  and  i,  2,  (3  insulated)  = 
2  (Ci  —  C«)  (C.  4-  ^  C.0 


G 


(75) 
(76) 

iyi) 


Capacitance  between  I,  S  and  2,  3  ::=  2  (Ci  +  Ca) (78) 

Capacitance  between  S  and  i,  2,  3  =  3  (C,  +  2C1,)  . . .   (79) 


It Q J 

FIG.    59 — DIMENSIO-VS   OF   A    SYMMETRICAL   THREE-PHASE   CABLE 

Cj  (76)  may  be  measured  in  the  ordinary  way, 
by  reading  the  throw  of  a  mirror  galvanometer  and 
comparing  with  the  throw  given  by  a  standard  con- 
denser. A  further  measurement  of  (78)  or  (79)  will 
give  a  simple  equation  to  find  C,2.  For  instance,  if 
measurements  were  taken  of  (78)  and  (79)  and  were 
found  to  be: — 


126 


CABLE  CHARACTERISTICS 


TABLE  XXVII— CAPACITANCE  AND  SUSCEPTANCE  PER  MILE  OF  THREE  CONDUCTOR 

PAPER  INSULATED  CABLES 


AREA 

IN 

CIRCULAR 

MILS 

B  &  S  NO. 

INSULATION  THICKNESS  IN  e4THS  OF  AN  INCH 

3     pv     3 
64    B^    64 

64    BY    ^ 

^    BY    ^ 

^    BY    ^ 

CAPACITANCE 

SUSCEPTANCE 
TO  NEUTRAL 

CAPACITANCE 

SUSCEPTANCE 
TO  NEUTRAL 

CAPACITANCE 

SUSCEPTANCE 
TO  NEUTRAL 

CAPACITANCE 

SUSCEPTANCE 
TO  NEUTRAL 

C 
1 

C 
12 

C 
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25 

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60 

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SUSCEPTANCE 
TO  NEUTRAL 

CAPACITANCE 

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CAPACITANCE 

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CAPACITANCE 

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SUSCEPTANCE 
TO  NEUTRAL 

CAPACITANCE 

SUSCEPTANCE 
TO  NEUTRAL 

CAPACITANCE 

SUSCEPTANCE 
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CAPACITANCE 

SUSCEPTANCE 
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1 

c 

12 

c 

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25 

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60 

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60 

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25 

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Capacitance— The  values  in  table  for  capacitance  were  derived  by  formulas  in  Alexander  Russel's  "Alternating  Currents." 
These  values  are  as  follows : — Cx  values  are  the  capacitance  in  microfarads  per  mile  between  one  conductor  and  the  other 
tv70  conductors  plus  sheath.  C1-2  values  are  the  mutual  capacitance  in  microfarads  per  mile  between  any  two  conductors. 
The  capacitance  to  neutral  is  twice  these  values.  C12  values  per  mile  are  used  in  the  application  of  Russel's  formulas  for 
determining  the  capacitance  corresponding  to  various  arrangements  of  the  three  conductors  and  the  sheath. 

The  Charging  Current  in  amperes  per  mile  for  each  conductor  to  neutral  =  susceptance  in  micromhos  to  neutral  (taken 
from  Table)  X  volts  to  neutral  X  10 ''. 

Dielectric  Constant — All  of  the  above  table  values  are  based  upon  a  value  for  the  dielectric  constant  K  of  3.5.  For  all 
other  values  of  K  the  table  values  will  change  in  direct  proportion.  Values  for  K  will  usually  be  found  between  the  follow- 
ing limits;  for  impregnated  paper  3.0  to  4.0;  for  varnished  cambric  4.0  to  6.0  and  for  rubber  4.0  to  9.0. 


CABLE  CHARACTERISTICS 


127 


TABLE  XXVIII— THREE-PHASE  CHARGING  KV-A  PER  MILE  OF  THREE-PHASE  CIRCUIT  Of 

THREE  CONDUCTOR  PAPER  INSULATED  CABLES 


25 

C  Y  C  L 

E  S 

AREA 

IN 

CIRCULAR 

MILS 

B  &  S  NO. 

CHARGING  KVA  PER  MILE  (EXPRESSED  IN  KV-A  3  PHASE)  FOR  PAPER  INSULATED  THREE 
CONDUCTOR  CABLES    BASED  UPON  A  VALUE  FOR'K'OF  3.5  AND  UPON  A  THICKNESS  OF 
INSULATION  SURROUNDING  THE  CONDUCTORS  AND  OF  THE  JACKET   INDICATED. 

220 
VOLTS 

■4 
64 

440 

VOLTS 

4 

64 

560 

VOLTS 

4 

64 

1100 

VOLTS 
-5. 
64 

2200 

VOLTS 

6 

64 

4400 
VOLTS 

6000 
VOLTS 

6600 
VOLTS 

6900 

VOLTS 

10 

63 

^ 

14 

64 

10 
64 

14 
64 

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13.200 
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16.500 
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The  values  in  Table  XXVIII  are  based  upon  a  value  for  the  dieleclric  constant  K  of  3.5.  For  all  other  values  of  K  the 
table  values  will  change  in  direct  proportion.  Values  for  K  will  usually  be  found  between  the  following  limits;  for  impreg- 
nated paper  3.0  to  4.0;  for  varnished  cambric  40  to  6.0  and  for  rubber  4.0  to  9.0. 


2  Ci-j-  2  Ca  =  0.410  mf.  per  mile (78) 

And  5  C,  +  6  C12  =  0.450  mf.  per  mile (79) 

Therefore  d   =  o.z6    mf.  per  mile 
C,2  =:  — 0.055  mf  per  mile 

Numerical  Examples — From  Table  XXVII  for  a 
250000  circ.  mil.,  three-conductor  cable  having  a  band 
of  insulation  surrounding  each  conductor  of  16/64  of 
an  inch  and  an  insulation  jacket  surorunding  all  three 
conductors  of  the  same  thickness,  the  following  values 
are  obtained : — 


Cx   =  0.260  mf.  per  mile. 

Cii  =  — 0.055  '"/.  /"'''  "I'l^^- 
Then,  in  the  order  in  which  the  capacitance  increases, — 

Capacitance  between  I  and  2  =  0.157  »'/■  per  mile (72) 

Capacitance  between  i  and  2,  3  :=  0.210  mf.  per  mile. .   (73) 
Capacitance  betv.een  I  and  S  (2  and  3  insulated)  = 

0.230  mf.  per  mile (74) 

Capacitance  between  I  and  S,  2  (3  insulated)  =  0.248 

mf.  per  mile  (75) 

Capacitance  between  I  and  S,  2,  3  ^  0.260  mf.  per 

mile (76) 

Capacitance  between  S  and  I,  2  (3  insluated)  =  0.363 

mf.  per  mite   (77) 


128 


CABLE  CHARACTERISTICS 


Capacitance  between  I,  S  and  3,  3  =  0.410  mf.  per 

mile (78) 

Capacitance  between  S  and  i,  2,  3  =:  0.450  mf.  per 

mile (79) 

COMPARISON   OF   CALCULATED   CAPACITANCE   WITH 
TEST  RESULTS 

The  difference  between  measured  results  of  capaci- 
tance and  the  results  calculated  by  the  above  formulas 
are  given  in  Fig.  60.  It  will  be  seen  that  in  all  cases 
these  calculated  results  are  less  than  the  corresponding 
test  results,  the  discrepancy  being  greater  as  the  con- 
ductor becomes  larger  and  the  separation  less.  The  dif- 
ferences vary  from  zero  to  as  much  as  eleven  percent 
for  the  largest  cable,  at  the  minimum  spacing  shown. 
The  discrepancy  is  greatest  with  the  minimum  thickness 
of  insulation.  Since  such  cables  would  be  used  only 
for  low-voltage  service,  the  charging  current  would  be 
small  and  consequently  this  error  would  probably  be  of 
little  importance.  For  6600  volt  cables  the  results  by 
the  formula  would  seem  to  be  approximately  five  per- 
cent too  low. 


4    6 


6     7      8     9     10  I  r    12    13    14    15    16    17    18   19    20 
THICKNESS  OF  INSULATION  SURROUNDING  CONDUCTOR  AND  OF 
JACKET  IN  64THS  OF  AN  INCH 

FIG.  60 — COMPARISON  OF  CALCULATED  AND  MEASURED  CAPACITANCES 

Tests  made  on  three  conductor  paper  insulated  cables,  K  =  3.5. 

The  cause  of  the  discrepancy  between  the  formula 
and  test  results  is  as  follows: — In  order  to  obtain  a 
mathematical  solution,  Russell  found  it  necessary  to 
make  certain  approximations  to  the  true  physical  condi- 
tions. Thus  the  resulting  mathematical  formula  can- 
not give  exact  results.  The  approximation  made  by 
Russell  is  very  close  to  the  actual  physical  fact  where 
the  conductors  are  small  compared  with  the  insulation 
thickness,  but  it  is  not  very  close  where  the  conductors 
are  large  compared  with  the  insulation. 

CHARGING  KV-A TABLE  XXVIII 

Table  XXVIII  contains  values  for  charging  cur- 
rent (expressed  in  kv-a,  three-phase)  for  three-conduc- 
tor paper  insulated  cables,  both  25  and  60  cycles,  based 
upon  a  value  for  K  of  3.5.  For  other  values  of  K,  the 
table  values  would  vary  in  proportion.  For  other 
thicknesses  of  insulation,  the  kv-a  values  would  vary  as 
the  susceptance  values  corresponding  to  the  thickness  of 
insulation  (See  Table  XXVII).  In  some  cases,  such 
for  instance,  as  grounded  neutral  systems,  the  thickness 


of  insulation  of  the  jacket  may  be  less  than  that  sur- 
rounding the  conductors.  In  such  cases  it  might  be  de- 
sirable to  calculate  the  susceptance  and  charging  cur- 
rent, if  accurate  results  were  desired.  The  values  for 
charging  current  corresponding  to  two  thicknesses  of 
insulation  are  included  for  some  of  the  commonly  em- 
ployed transmission  voltages. 

These  kv-a  values  were  calculated  by  using  the 
values  for  susceptance  in  Table  XXVII  which,  in  turn, 
were  derived  from  the  capacitance  in  the  same  table  ob- 
tained by  formulae  (70)  and  (71).  Thus  a  350000 
circ.  mil  cable  with  10/64  and  10/64  paper  insulation 
has  a  60  cycle  susceptance  to  neutral  of  167  micromhos 
per  mile.  Since  the  charging  current  in  amperes  to 
neutral  equals  the  susceptance  to  neutral  X  volts  to 
neutral  X  lo"*  and  assuming  6600  volts,  three-phase 
between  conductors,  we  have: — 

6600 

167  X  -^-rr^X  10-*  =  0.637  amperes  to  neutral. 

Charging  kv-a  —  0.637  X  3815  X  3  =  7-25  kv-a, 
as  indicated  in  Table  XXVIII. 

VALUES   FOR   K 

The  capacitance  of  any  cable  depends  upon  the  di- 
electric constant  of  the  insulating  material  and  a  dimen- 
sion term  or  form  factor.  The  dielectric  constant 
should  be  determined  from  actual  cables  and  not  from 
samples  of  material.  The  usual  range  in  value  for  K 
is  given  below. 

Value  of  K 

Impregnated  Paper 3.0  to  4.0 

Varnish  Cambric  4.0  to  6.0 

Rubber 4.0  to  9.0 

All  values  in  Tables  XXVII  and  XXVIII  are  based 
upon  a  value  of  K  of  3.5.  For  all  other  values  of  K 
all  table  values  will  vary  in  the  same  proportion  as  their 
K  values.  The  actual  value  of  permittivity  of  most 
paper  insulation  runs  about  ten  percent  less  than  the 
value  3.5  which  has  been  used  in  calculating  the  accom- 
panying table  values.  The  true  alternating-current  ca- 
pacitance is  always  considerably  lower  than  the  capaci- 
tance measured  with  ballistic  galvanometer. 

REFERENCES 

"Electric  Power  Conductors,"  by  W.  S.  Del  Mar. 
"Electric  Cables,"  by  Coylc  and  Howe,  London,  England. 
"The  Heating  of  Cables  with  Current,"  by  Melsoni  &  Booth. 
Journal  I.  E.  E.,  Vol.  47 — 191 1. 

"Current  and  Rating  of  Electric  Cables,"  by  Atkinson  S.- 
Fisher, Trans.  A.  I.  E.  E.,  1913,  p.  325. 

"The  Heating  of  Cables  Carrying  Current,"  bv  Dushman. 
Trans.  A.  I.  E.  E.,  1913,  p.  333. 

"Effect  of  Moisture  in  the  Earth  on  Temperature  of  Under- 
ground Cables,"  by  Imlay,  Trans.  A.  I.  E.  E.,  1915,  Part  I,  p. 
2^3- 

"Temperature  Rise  of  Insulated  Lead  Covered  Cables,"  by 
Richard  A.  Powell,  Trans.  A.  I.  E.  E.,  1916,  Part  II,  p.  1017. 

"The  Restoration  of  Service  After  a  Necessary  Interrup- 
tion," by  Rickets,  Trans.  A.  I.  E.  E.,  1916,  Part  II,  p.  635. 

"The  Dielectric  Field  in  an  Electric  Power  Cable,"  by 
Atkinson,  Trans.  A.  I.  E.  E.,  June  1919. 

"The  Current  Carrying  Capacity  of  Lead  Covered  Cables," 
by  Atkinson,  Journal  A.  I.  E.  E.,  Sept.  1920. 


CHAPTER  XIV 

SYNCHRONOUS  MOTORS  AND  CONDENSERS  FOR 

POWER-FACTOR  IMPROVEMENT 


BEFORE  discussing  the  employment  of  syn- 
chronous machinery  for  improving  the  power- 
factor  of  circuits,  it  may  be  desirable  to  review 
how  a  change  in  power-factor  affects  the  generators 
supplying  the  current. 

Fig.  6i  shows  the  effect  of  in-phase,  lagging  and 
leading  components  of  armature  current  upon  the  field 
strength  of  generators*.  A  single-coil  armature  is  il- 
lustrated as  revolving  between  the  north  and  south 
poles  of  a  bipolar  alternator.  The  coil  is  shown  in 
four  positions  90  degrees  apart,  corresponding  to  one 
complete  revolution  of  the  armature  coil.  The  direc- 
tion of  the  field  flux  is  assumed  to  be  constant  as  in- 
dicated by  the  arrows  on  the  field  poles  of  each  illus- 
tration. In  addition  to  this  field  flux,  when  current 
flows  through  the  armature  coil  another  magnetic  flux 
is  set  up,  magnetizing  the  iron  in  the  armature  in  a  di- 
rection at  right  angles  to  the  plane  of  the  armature  coil. 
This  will  be  referred  to  as  armature  flux. 

This  armature  flux  varies  with  the  armature  cur- 
rent, being  zero  in  a  single-phase  generator  when  no 
armature  current  flows,  and  reaching  a  maximum 
when  full  armature  current  flows.  It  changes  in  direc- 
tion relative  to  the  field  flux  as  the  phase  angle  of  the 
armature  current  changes. 

The  revolving  armature  coil  generates  an  alternat- 
ing voltage  the  graph  of  which  follows  closely  a  sine 
wave,  as  shown  in  Fig.  61.  When  it  occupies  a  verti- 
cal plane  marked  start  no  voltage  is  generated,  for  the 
reason  that  the  instantaneous  travel  of  the  coil,  is 
parallel  with  the  field  flux.**  As  the  coil  moves  for- 
ward in  a  clockwise  direction,  the  field  enclosed  by 
the  armature  coil  decreases;  at  first  slowly  but  then 
more  rapidly  until  the  rate  of  change  of  flux  through 
the  coil  becomes  a  maximum  when  the  coil  has  turned 
90  degrees,  at  which  instant  the  voltage  generated  be- 
comes a  maximum.  As  the  horizontal  position  is  passed 
the  voltage  decreases  until  it  again  reaches  zero  when 
the  coil  has  traveled  180  degrees  or  occupies  again  a 
vertical  plane.  As  the  travel  continues  the  voltage 
again  starts  to  increase  but  since  the  motion  of  the  coil 


♦For  a  more  detailed  discussion  of  this  subject  the  reader 
is  referred  to  excellent  articles  by  F.  D.  Newbury  in  the 
Electric  Journal  of  April  1918,  "Armature  Reaction  of  Poly- 
phase Alternators" ;  and  of  July  1918.  "Variation  of  Alternator 
Excitation  with  Load". 

**For  the  sake  of  simplicity  this  and  the  following  state- 
ments are  based  upon  the  assumption  that  armature  reaction 
does  not  shift  the  position  of  the  field  flux.  Actually,  under 
load,  the  armature  reaction  causes  the  position  of  the  field  flux 
to  be  shifted  toward  one  of  the  pole  tips,  so  that  the  position 
of  the  armature  coil  is  not  quite  vertical  at  the  instant  of  zero 
voltage  in  the  coil. 


relative  to  the  fixed  magnetic  field  is  reversed  the  volt- 
age in  the  coil  builds  up  in  the  reverse  direction  dur- 
ing the  second  half  of  the  revolution.  When  the  coil 
has  reached  the  two  270  degree  position  the  voltage 
has  again  become  maximum  but  in  the  opposite  direc- 
tion to  that  when  the  coil  occupied  the  position  of 
90  degrees.  When  the  coil  returns  to  its  original  posi- 
tion at  the  start  the  voltage  has  again  dropped  to  zero, 
thus  completing  one  cycle.  » 

If  the  current  flowing  through  this  armature  coil 
is  in  phase  with  the  voltage,  it  will  produce  cross  mag- 
netization in  the  armature  core,  in  a  vertical  direction, 
as  indicated  by  the  arrows  at  the  90  and  270  degree 
positions.  The  cross  magnetization  neither  opposes 
nor  adds  to  the  field  flux  at  low  loads  and  therefore  has 
comparatively  little  influence  on  the  field  flux.  At 
heavy  loads,  however,  this  cross  magnetization  has  con- 
siderable demagnetizing  effect,  due  to  the  shift  in  ro- 
tor position  resulting  from  the  shifting  of  the  field  flux 
at  heavy  loads. 

If  the  armature  is  carrying  lagging  current,  this 
current  will  tend  to  magnetize  the  armature  core  in 
such  a  direction  as  to  oppose  the  field  flux.  This  ac- 
tion is  shown  by  the  middle  row  of  illustrations  of 
Fig.  61.  Under  these  illustrations  is  shown  a  current 
wave  lagging  90  degrees  representing  the  component 
of  current  required  to  magnetize  transformers,  induc- 
tion motors,  etc.  When  the  lagging  component  of  cur- 
rent reaches  its  maximum  value  the  armature  coil  will 
occupy  a  vertical  position  (position  marked  start,  180 
degrees  and  360  degrees)  and  in  this  position  the  arma- 
ture flux  will  directly  oppose  the  field  flux,  as  indicated 
by  the  arrows.  The  result  is  to  reduce  the  flux  thread- 
ing the  armature  coil  and  thus  cause  a  lowering  of  the 
voltage.  This  lagging  current  encounters  resistance 
and  a  relatively  much  greater  reactance,  each  of  which 
consumes  a  component  of  the  induced  voltage,  as 
shown  in  Fig.  62.  When  the  armature  current  is  lag- 
ging, the  voltage  induced  by  armature  inductance  is  in 
such  a  direction  as  to  subtract  from  the  induced  volt- 
age, and  thus  the  voltage  is  still  further  lowered,  as  a 
result  of  the  armature  self  induction.  In  order  to 
bring  the  voltage  back  to  its  normal  value  it  will  be 
necessary  to  increase  the  field  flux  by  increasing  the 
field  current.  Generators  are  now  usually  designed  of 
sufficient  field  capacity  to  compensate  for  lagging 
loads  of  80  per  cent  power-factor. 

If  the  armature  is  carrying  a  leading  current  this 
leading  component  will  tend  to  magnetize  the  armature 
core  in  such  a   direction  as  to  add  to  the  field  flux. 


130 


SYNCHRONOUS  MOTORS  AND  CONDENSERS    FOR    POWER-FACTOR    CORRECTION 


This  action  is  shown  by  the  bottom  row  of  illustrations 
of  Fig.  6i.  Under  these  illustrations  is  shown  a  current 
wave  leading  the  voltage  wave  by  90  degrees.  When 
the  leading  component  of  current  reaches  its  maximum 
values,  the  armature  coil  will  again  occupy  vertical 
positions,  but  the  armature  flux  will  add  to  that  of  the 
field  flux,  as  indicated  by  the  arrow.  The  resulting 
flux  threading  the  armature  coil  is  thus  increased  caus- 
ing a  rise  in  voltage.  This  leading  current  flowing 
through  the  generator  armature  encounters  resistance 
and  a  relatively  much  greater  reactance,  each  of  which 
consumes  a  component  of  the  induced  voltage,  as 
shown  in  Fig.  62.     When  the  armature  current  is  lead- 

OROSS  MAQNETIZINQ  EFFECT  OF    IN-PHASE  ARMATURE  CURRENT 

laUCH  U  OONSUMCO  ST  INOANOESCCNT  LAMPt' 


KM  5iB  53  53  EiS 


SUSTRACTIVE  EFFECT  OF  A  LAGGING  ARMATURE  CURRENT 

'■UCM  M  THE  MAONrtlliNa  CUBOCMT  Of   TDAMftTOOMIOS  AND  INDUCTtOH  HOTOMi 

Eia  g'S  K:3  5  3   E:S 


ADDITIVE  EFFECT  OF  A  LEADING  ARMATURE  CURRENT 

itUCH    U  THI    CHUtOlNO  OunnCNT    OF    TRANSMISSION    UNU  (J*    OVM    I>OITEX>  STN     MOTDBS 

I  I  I 


KiS  5S  E:B  BS   E:S 


FIG.  61- 


-EFFECT  OF  ARMATURE  CURRENT  UPON  FIELD  EXCITATION  OF 
ALTERNATING-CURRENT  GENERATORS 


ing,  the  voltage  induced  by  armature  inductance  is  in 
such  a  direction  as  to  add  to  the  induced  voltage  und 
thus  the  voltage  at  the  alternator  terminals  is  .still 
further  increased  as  the  result  of  armature  self-induc- 
tion. In  order  to  reduce  the  voltage  to  its  normal 
value  it  is  necessary  to  decrease  the  field  flux  by  de- 
creasing the  field  current. 

With  alternators  of  high  reaction  the  magnetizing 
or  de-magnetizing  effect  of  leading  or  lagging  current 
will  be  greater  than  in  cases  where  the  armature  reac- 
tion is  low.  For  instance  if  the  alternator  is  so  de- 
signed that  the  ampere  turns  of  the  armature  at  full 
armature  current  are  small  compared  to  its  field  am- 
pere turns,  the  voltage  of  such  a  machine  would  be  less 
disturbed  with  a  change  in  power-factor  of  the  arma- 


ture current  than  in  an  alternator  having  armature 
ampere  turns  large  compared  with  its  field  ampere 
turns. 

Modern  alternators  are  of  such  design  that  when 
carrying  rated  lagging  current  at  zero  power-factor 
they  require  approximately  200  to  250  percent  of  their 
no-load  field-current  and  when  carrying  rated  leading 
current  at  zero  power-factor  they  require  approximate- 
ly — 15  to  -|-I5  percent  of  their  no-load  field  current. 
Thus  with  lagging  armature  current  the  iron  will  be 
worked  at  a  considerable  higher  point  on  the  satura- 
tion curve  and  the  heating  of  the  field  coils  will  in- 
crease because  of  the  greater  field  current  required. 

The  voltage  diagrams  of  Fig.  62  are  intended  to 
show  only  the  effect  of  armature  resistance  and  arma- 
ture reactance  upon  voltage  variation.     Voltage  regu- 


FIG.    62 — VECTORS    ILLUSTRATING   THE   EFFECT   OF   ARMATURE    REACT- 
ANCE AND  RESISTANCE  UPON  THE  TERMINAL  VOLTAGE  FOR  IN-PHASE, 
LEADING  AND  LAGGING  CURRENTS 

lation  is  the  combined  effect  of  armature  impedance 
and  armature  reaction.  Turbogenerators  have,  for 
instance,  very  low  armature  reactance  but  their  arma- 
ture reaction  is  higher,  so  that  the  resulting  voltage 
regulation  may  not  be  materially  different  from  that  of 
a  machine  with  double  the  armature  reactance. 
Under  normal  operation  armature  reaction  is  a  more 
potent  factor  in  determining  the  characteristics  of  a 
generator  than  armature  reactance.  In  the  case  of  a 
generator  with  a  short  circuit  ratio  of  unity,  this  total 
reactive  effect  may  be  due,  15  percent  to  armature  re- 
actance and  85  percent  to  armature  reaction. 

For  the  case  illustrated  by  V\g.  62  the  field  flux 
corresponds  to  the  induced  voltage  indicated,  but  the 
field  current  does  not.  The  field  current  corresponds 
to  a  value  obtained  by  substituting  the  full  synchron- 
ous impedance  drop  for  that  indicated. 


SyNCHKONOUS  MOTORS  AND  CONDENSEKS  I'OK  POWER-FACTOR  CORRECTION 


I3« 


SYNCHRONOUS  CONDENSERS  AND  I'HASE   MODIFIERS 

The  term  "synchronous  condenser"  applies  to  a 
synchronous  machine  for  raising  the  power- factor  of 
circuits.  It  is  simply  floated  on  the  circuit  with  its 
fields  over  excited  so  as  to  introduce  into  the  circuit  a 
leading  current.  Such  machines  are  usually  not 
intended  to  carry  a  mechanical  load.  When  this  dou- 
ble duty  is  required  they  are  referred  to  as  synchron- 
ous motors  for  operation  at  leading  power-factor. 
On  long  transmission  circuits,  where  synchronous  con- 
densers are  used  in  parallel  with  the  load  for  varying 
the  power-factor,  thereby  controlling  the  transmission 
voltage,  it  is  sometimes  necessary  to  operate  them  with 
under  excited  fields  at  periods  of  lightloads.  They  are 
then  no  longer  synchronous  condensers  but  strictly 
speaking  become  synchronous  reactors. 

Whether  synchronous  motors  for  operation  at 
leading  power-factor,  synchronous  condensers  or  syn- 
chronous reactors  be  used  they  virtually  do  the  same 
thing,  that  is;  their  function  is  to  change  the  power- 
factor  of  the  load  by  changing  the  phase  angle  between 
the  armature  current  and  the  terminal  voltage.     They 

TABLE  R— SYNCHRONOUS  CONDENSER  LOSSES 


Kv-a 

Loss  (Kw) 

Kv-a 

Loss  (Kw) 

100 

12 

3500 

180 

200 

18 

5000 

220 

300 

22 

7500 

320 

500 

3-' 

lOOOO 

420 

750 

47 

15000 

620 

1000 

55 

20000 

820 

1500 

70 

25000 

1000 

2000 

IJO 

35000 

1400 

2500 

130 

50000 

2000 

are,  therefore,  sometimes  referred  to  as  "phase  modi- 
fiers." This  latter  name  seems  more  appropriate  when 
the  machine  is  to  be  operated  both  leading  and  lagging, 
as  when  used  for  voltage  control  of  long  transmission 
lines. 

Rating  —  Sjmchronous  condensers  as  regularly 
built  may  be  operated  at  from  30  to  40  percent  of  their 
rating  lagging,  depending  upon  the  individual  design. 
Larger  lagging  loads  result  in  unstable  operation  on 
account  of  the  weakened  field.  Phase  modifiers  can 
be  designed  to  operate  at  full  rating,  both  leading  and 
lagging,  but  they  are  larger,  require  larger  exciters, 
have  a  greater  loss  and  cost  15  to  20  percent  more 
than  standard  condensers. 

Starting — Condensers  are  furnished  with  squir- 
rel-cage damper  windings,  to  prevent  hunting,  which 
also  provides  a  starting  torque  of  approximately  30 
percent  of  normal  running  torque.  They  have  a  pull- 
in  torque  of  around  15  percent  of  running  torque. 
The  line  current  at  starting  varies  from  50  to  100  per- 
cent of  normal.  The  larger  units  are  sometimes 
equipped  for  forced  oil  lubrication,  which  raises  the 
rotor  sufficiently  to  permit  of  oil  entering  the  bearing, 
thus  reducing  the  starting  current. 


Mechanical  Load — Synchronous  condensers  are 
generally  built  for  high  speeds  and  equipped  with 
shafts  of  small  diameter.  If  they  are  to  be  used  to 
transmit  some  mechanical  power  it  may  be  necessary 
to  equip  them  with  larger  shafts  and  bearings,  particu- 
larly if  belted  rather  than  direct  connected.  If  a 
phase  modifier  is  to  furnish  mechanical  energy  and  at 
the  same  time  to  operate  lagging  at  times  of  light  load 
for  the  purpose  of  holding  down  the  voltage  on  an  un- 
loaded transmission  line  there  may  be  danger  of  the 
machine  falling  out  of  step,  if  a  heavy  mechanical  load 
occurs  when  the  machine  is  operating  with  a  weak 
field. 

Losses — At  rated  full  load  leading  power-factor 
the  total  losses,  including  those  of  the  exciter,  will  vary 
from  approximately  12  percent  for  the  smallest  capaci- 
ty to  approximately  four  percent  for  the  larger  capaci- 
ty 60  cycle  synchronous  condensers.     The  approximate 


76  1 00  1 26 

FIELD  AMPERES 

FIG.  63— V-CUKVES  OF  A  PHASE  MODIFIER 

values  given  in  Table  R  may  be  of  service  for  prelimin- 
ary purposes. 

"V"  Curves — The  familiar  V  curves  shown  in 
Fig.  63  serve  to  give  some  idea  of  the  variation  in  field 
current  for  a  certain  phase  modifier  when  operating 
between  full  load  lagging  and  full  load  leading  kv-a.* 
For  this  particular  machine  the  excitation  must  be  in- 
creased from  112  amperes  at  no  load  minimum  input 
or  unity  power-factor  to  155  amperes  at  full  kv-a  out- 
put leading  or  a  range  of  1.4  to  i  in.  field  excitation. 
For  operation  between  full  lagging  and  full  leading, 
with  no  mechanical  work  done,  the  range  of  excitation 
is  from  67  to  155  or  2.3  to  i. 

Generators  as  Condensers — Ordinary  alternators 
may  be  employed  as  synchronous  condensers  or  syn- 
chronous motors  by  making  proper  changes  in  their 
field  poles  and  windings  to   render  them   self-starting 


♦These  curves  have  been  reproduced  from  H.  B.  Dwight's 
book  "Constant  Voltage  Transmission". 


13^ 


SYNCHRONOUS  MOTORS  AND  CONDENSERS    FOR    POWRR-fACTOR    CORRECTION 


and  safely  insulated  against  voltages  induced  in  the 
field  when  starting. 

Where  transmission  lines  feed  into  a  city  net  work 
and  a  steam  turbine  generator  station  is  available  these 
generating  units  can  serve  as  synchronous  condensers 
by  supplying  just  enough  steam  to  supply  their  losses 
and  keep  the  turbine  cool.  When  operated  in  this  way 
they  make  a  reliable  standby  to  take  the  important  load 
quickly  in  case  of  trouble  on  a  transmiss>on  line. 

Location  for  Condensers — The  nearer  the  center 
of  load  that  the  improvement  in  power-factor  is  made 
the  better,  as  thereby  the  greatest  gain  in  regulation,, 
greatest  saving  in  conductors  and  apparatus  are  made 
since  distribution  lines,  transformers,  transmission 
lines  and  generators  will  all  be  benefited. 

How  High  to  Raise  the  Power-Factor — Theoreti- 
cally for  most  efficient  results  the  system  power 
factor  should  approach  unity.  The  cost  of  synchron- 
ous apparatus  having  sufficient  leading  current  capaci- 
ty to  raise  the  power-factor  to  unity  increases  so 
rapidly  as  unity  is  approached,  as  to  make  it  unecono- 
mical to  carry  the  power-factor  correction  too  high. 
Not  only  the  cost  but  also  the  power  loss  chargeable  to 
power-factor  improvement  mounts  rapidly  as  higher 
power-factors  are  reached.  This  is  for  the  reason  that 
the  reactive  kv-a  in  the  load  corresponding  to  each  per- 
cent change  in  power-factor  is  a  maximum  for  power- 
factors  near  unity.  It  usually  works  out  that  it 
doesn't  pay  to  raise  the  power  factor  above  90  to  95 
percent,  except  in  cases  where  the  condenser  is  used 
for  voltage  control,  rather  than  power-factor  improve- 
ment. 

DETERMINING   THE    CAPACITY   OF   SYNCHRONOUS    MOTORS 
AND  CONDENSERS  FOR  POWER-FACTOR  IMPROVEMENT 

A  very  simple  and  practical  method  for  determining 
the  capacity  of  synchronous  condensers  to  improve  the 
power-factor  is  by  aid  of  cross  section  paper.  A  very 
desirable  paper  is  ruled  in  inch  squares,  sub- ruled  into 
10  equal  divisions.  With  such  paper,  no  other  equip- 
ment is  required. 

With  a  vector  diagram  it  is  .astonishing  how  easy 
it  is  to  demonstrate  on  cross  section  paper,  the  effect 
of  any  change  in  the  circuit.  A  few  typical  cases  are 
indicated  in  Fig.  64.  These  diagrams  are  all  based  up- 
on an  original  circuit  of  3000  kv-a  at  70  percent  power- 
factor  lagging,  shown  by  (i).  It  is  laid  off  on  the 
cross  section  paper  as  follows.  The  power  of  the  cir- 
cuit is  70  percent  of  3000  or  2100  kw,  which  is  laid  off 
on  line  AB,  by  counting  21  sub-divisions,  making 
each  sub-division  represent  100  kw  or  100  kv-a.  Now 
lay  a  strip  of  blank  paper  over  the  cross  section  paper 
and  make  two  marks  on  one  edge  spaced  30  sub-divi- 
sions apart.  This  will  then  be  the  length  of  the  line  AC. 
This  blank  sheet  is  now  laid  over  the  cross  section 
paper  with  one  of  the  marks  at  the  edge  held  at  the 
point  A.  ■'The  other  end  of  the  paper  is  moved  down- 
ward until  the  second  mark  falls  directly  below  the 
point  B    thus   locating   point  C.     The   length    of   the 


line  BC  represents  the  lagging  reactive  kv-a  in  the  cir- 
cuit, in  this  case  2140  kv-a. 

Diagram  (2)  shows  the  effect  of  adding  a  1500 
kv-a  synchronous  condenser  to  the  original  circuit. 
The  full  load  loss  of  this  condenser  is  assumed  as  70 
kw.  The  resulting  kv-a  and  power- factor  are  de- 
termined as  follows:  Starting  from  the  pomt  C  trace 
to  the  right  a  line  0.7  of  a  division  long.  This  is 
parallel  to  the  line  AB  for  the  reason  that  it  is  true 
power,  so  that  there  is  now  2170  kw  true  energy.  The 
black  triangle  represents  the  condenser,  the  line  CD, 
15  divisions  long,  representing  the  rating  of  the  con- 
denser. In  this  case,  however,  the  vertical  line  is 
traced  upward  in  place  of  downward,  because  the  con- 
denser kv-a  is  leading.  This  condenser  results  in  de- 
creasing the  load  from  3000  kv-a  at  70  percent  power- 
factor  to  2275  kv-a  at  95.4  percent  power-factor.  The 
line  AD  represents  in  magnitude  and  direction,  the  re- 
sulting kv-a  in  this  circuit.  The  power-factor  of  the 
resulting  circuit  is  the  ratio  of  the  true  energy  in  kw  to 
the  kv-a  or  95.4  percent,  in  this  case.  Since  the  line 
AD  lays  below  the  line  AB,  that  is  in  the  lagging  direc- 
tion, the  power- factor  is  lagging. 

Diagram  (3)  is  the  same  as  (2)  except  that  the 
condenser  is  larger,  being  just  large  enough  to  neu- 
tralize all  of  the  lagging  component  of  the  load,  result- 
ing in  a  final  load  of  2215  kw  at  100  percent  power-fac- 
tor. Diagram  (4)  is  similar  to  (3)  except  that  a 
still  larger  condenser  is  shown.  This  condenser  not 
only  neutralizes  all  of  the  lagging  kv-a  of  the  load  but 
in  addition  introduces  sufficient  leading  kv-a  into  the 
circuit  to  give  a  leading  resultant  power-factor  of  91 
percent  with  an  increase  in  kv-a  of  the  resulting  cir- 
cuit from  2215  of  (3)  to  2400  kv-a  of  (4). 

Diagram  (5)  illustrates  the  addition  to  the  original 
chxuit  of  a  100  percent  power-factor  synchronous 
motor  of  600  hp.  rating  As  this  motor  has  no  leading 
or  lagging  component,  there  is  no  vertical  projection. 
The  power- factor  of  the  circuit  is  raised  from  70  to 
y7  percent  as  the  result  of  the  addition  of  500  kw  true 
power  (load  plus  loss  in  motor)  to  the  circuit.  A  re- 
sistance load  would  have  this  same  effect. 

Diagram  (6)  shows  a  450  kw  (600  hp.)  syn- 
chronous motor  of  625  kv-a  input  at  80  percent  lead-j 
ing  power-factor  added  to  the  original  circuit.  The 
input  to  this  motor  (including  losses)  is  assumed  to  be 
500  kw.  The  resulting  load  for  the  circuit  is  3150 
kv-a  at  82.5  percent  lagging  power-factor. 

The  Diagram  (7)  shows  an  850  kw,  (1140  hp.) 
synchronous  motor  generator  of  1666  kv-a  input  at  60 
percent  power-factor  leading  added  to  the  original  cir- 
cuit. This  gives  a  resulting  load  of  3200  kv-a  at  96.9 
percent  lagging  power- factor. 

Diagram  (8)  shows  the  addition  to  the  original 
circuit  of  the  following  loads,  including  losses. 

A  550  kw  synchronous  converter  at  100  percent  power- 
factor. 

A  650  kw  inljction  motor  at  70  percent  lagging 
power-factor. 

.\     500  kw  '\  iirhronoiis  mi>tor. 


SYNCHRONOUS  MOTORS  AND  CONDENSERS    FOR    POWER-FACTOR    CORRECTION 


133 


The  resultant  load  of  this  circuit  is  3800  kw,  and 
if  a  power-factor  of  95  percent  lagging  is  desired  the 
total  kv-a  will  be  4000.  The  line  AD  may  be  located 
by  a  piece  of  marked  paper  and  the  capacity  of  the 
necessary  synchronous  motor  scaled  oflf.  This  is 
found  to  be  1650  kv-a  at  30.3  percent  power-factor. 


The  Circle  Diagram — The  circle  diagram  in  Fig.  65 
shows  the  fundamental  relations  between  true  kw,  reac- 
tive kv-a  and  apparent  kv-a  corresponding  to  different 
power-factors,  the  values  upon  the  chart  being  read  to 
any  desired  scale  to  suit  the  numerical  values  of  the 
problem  under  consideration.     This   diagram   is   suffi- 


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FIG.  64— EXAMPLES  IN  POWER-FACTOR  IMPROVEMENT 


134 


SYNCHRONOUS  MOTORS  AND  CONDENSERS    FOR    POIVER-EACTOR    CORRECTION 


ciently  accurate  for  ordinary  power-factor  problems. 
In  place  of  drawing  out  the  vector  diagrams  as  just 
explained  they  are  traced  out  with  a  pin  point  on  the 
circle  diagram. 

Assume  again  a  load  of  2100  kw  at  70  percent 
power-factor  lagging,  and  that  the  power-factor  is  to 
be  raised  to  95.4  percent  as  in  (2)  of  Fig.  64,  and  that 
the  loss  in  the  condenser  necessary  to  accomplish  this 
is  again  taken  as  70  kw.  The  capacity  of  the  syn- 
chronous condenser  may  be  traced  on  the  circle  dia- 
gram as  follows:  From  the  true  power  load  of  2100 
kw  (top  horizontal  line)    follow  vertically  downward 


of  the  condenser  would  be  the  hypotenuse  rather  than 
the  vertical  projection.  The  error  in  assuming  the 
vertical  projection  as  the  rating  of  the  condenser  is 
negligible  unless  the  condenser  furnishes  mechanical 
power,  in  which  case  the  hypotenuse  should  be  marked 
on  a  separate  strip  of  paper  and  its  length  determined 
from  the  kv-a  scale. 

ADVANTAGE    OF    HIGH    POWER-FACTOR 

Less  Capacity  Installed — Low  power-factors  de- 
mand larger  generators,  exciters,  transformers,  switch- 
ing equipment  and  conductors.     Loads   of   70  percent 


1000 


2000 


3000 


POWER  KW 
4000  5000 


6000 


7000 


8000 


9000 


10.000 


%  POWER-FACTOR 
FIG.  65 — RELATION  BETWEEN  ENERGY  LOAD,  APPARENT  LOAD  AND  REACTIVE  KV-A  FOR  DIFFERENT  POWER  FACTORS 


until  the  diagonal  line  representing  70  percent  power- 
factor  is  reached.  This  is  opposite  2140  kv-a  reactive 
component.  From  the  point  thus  obtained,  go  hori- 
zontally to  the  right  a  distance  representing  70  kw 
power.  From  this  point  go  vertically  upward  until 
the  diagonal  line  representing  95.4  percent  power-fac- 
tor is  reached.  Then  read  the  amount  of  reactive  kv-a 
(640)  corresponding  to  this  last  point.  The  original 
lagging  component  of  2140 — 640=^1500  kv-a  which  is 
approximately  the  capacity  of  the  condenser  necessary 
to  accomplish  the  above  results.     Actually  the  rating 


power-factor  demand  equipment  of  28  percent  greater 
capacity  than  would  be  required  if  the  power-factor 
were  90  percent.  The  cost  of  apparatus  for  opera- 
tion at  70  percent  power-factor  would  be  approxi- 
mately 15  percent  greater  than  the  cost  of  similar  ap- 
paratus for  90  percent  power-factor  operation,  since 
the  capacity  of  apparatus  to  supply  a  certain  amount 
of  energy  is  inversely  proportional  to  the  power-factor. 
Higher  Efficiency — Assume  that  the  power-factor 
of  a  1000  kv-a  (700  kw  at  70  percent  power- factor) 
transmission  circuit  is  raised  to  90  percent.  As  the  cop- 


SYNCHRONOUS  MOTORS  AND  CONDENSERS    FOR    POWER-FACTOR    CORRECTION 


'35 


per  loss  varies  as  the  square  of  the  current,  raising  the 
power- factor  reduces  the  copper  loss  approximately 
40  percent.  If  we  assume  an  efficiency  for  the  genera- 
tor of  93  percent  (one  percent  copper  loss)  ;  for  com- 
bined raising  and  lowering  transformers  94  percent 
(three  percent  copper  loss)  and  for  the  transmission 
line  92  percent,  the  saving  in  copper  loss  correspond- 
ing to  90  percent  power-factor  operation  would  be  as 
follows : 

Generators 0.4  percent 

Transformers 1.2  percent 

Transmission  line  ....  3.2  percent 

Total  4.8  percent  or  approximately  33  kw. 

To  raise  the  power-factor  to  90  percent  would  re- 
quire a  synchronous  condenser  of  375  kv-a  capacity. 
This  size  condenser  would  have  a  total  loss  of  about 
30  kw,  resulting  in  a  net  gain  in  loss  reduction  of 
three  kw.  Against  this  gain  would  be  chargeable,  the 
interest  and  depreciation  of  the  condenser  cost  with  its 
accessories,  also  any  cost  of  attendance  which  there 
might  be  in  connection  with  its  operation.  It  is  evi- 
dent that  in  this  case  it  would  not  pay  to  install  a  con- 
denser if  increased  efficiency  were  the  only  motive. 

TABLE  S— COST  OF  POWER-FACTOR  CORRECTION 
WITH  SYNCHRONOUS  MOTORS 


Syn.  Motor 
Kv-a 

Motor  Will  Furnish 

Chargeable  to 
Power-Factor  Correction 

Mech. 
Kw 

Leading 
Kv-a 

Loss 
Kw 

Difference 
in  Price 

140 
280 
420 
700 
1050 
1400 

100 
200 
300 
500 
750 

1000 

100 
200 
300 
500 
750 
1000 

1.6 

2.5 
50 
8.0 
9.0 
14.0 

$500.00 

500.00 

500.00 

800.00 

1000.00 

1200.00 

The  improvement  in  power-factor  can  be  more 
cheaply  and  efficiently  obtained  by  the  installation  of 
one  or  more  synchronous  motors  designed  for  opera- 
tion at  leading  power-factor.  Sufficient  capacity  of 
these  will  give,  in  addition  to  mechanical  load,  suffi- 
cient leading  current  to  raise  the  power-factor  to  90 
percent.  The  extra  expense  and  increased  loss  of 
synchronous  motors  enough  larger  to  furnish  the  nec- 
essary leading  component  for  power-factor  correction 
is  very  small.  Table  S  gives  in  a  very  approximate 
way,  some  idea  of  the  amount  of  loss  and  proportional 
cost  of  synchronous  motors  chargeable  to  power-factor 
improvement  when  delivering  both  mechanical  power 
and  leading  current. 

Thus  if  a  synchronous  condenser  is  used  on  the 
above  circuit  there  is  a  loss  of  30  kw,  chargeable  to 
power-factor  improvement,  whereas  if  a  synchronous 
motor  of  sufficient  capacity  (530  kv-a)  to  give  375  kw 
mechanical  work  and  at  the  same  time  the  necessary 
375  kv-a  leading  current  for  power-factor  improve- 
ment, the  extra  loss  chargenble  to  power-factor  im- 
provement would  be  something  like  six  kw.  The  in- 
creased cost  of  a  synchronous  motor  to  furnish  375 
kv-a  leading  current  in  addition  to  375  kw  power 
would  be  about  $600  vi'hereas  the  cost  of  a  375  kv-a 


condenser  would  be  in  the  neighborhood  of  $4000. 
Varying  costs  and  designs  make  cost  and  loss  values 
unreliable.  They  are  given  here  only  to  illustrate  the 
points  which  should  be  considered  when  considering 
synchronous  motors  vs  synchronous  condensers. 

Improved  Voltage  Regulation — The  voltage  drop 
under  load  for  generators,  transformers  and  trans- 
mission lines  rapidly  increases  as  the  power-factor  goes 
down.  Table  T  gives  an  idea  of  the  variation  in  volt- 
age drop  corresponding  to  various  power-factors  at  60 
cycles. 

Automatic  voltage  regulation  may  be  used  to  hold 
the  voltage  constant  at  the  generators  or  at  some  other 
point,  but  it  cannot  prevent  voltage  changes  at  all 
points  of  the  system. 

Increased  Plant  Capacity — The  earlier  alternators 
were  designed  for  operation  at  100  percent  power-fac- 
tor with  prime  movers,  boilers,  etc  installed  on  the 
same  basis.  Increasing  induction  motor  loads  have 
resulted  in  power-factors  of  70  and  80  percent.  As  a 
result,  some  of  the  older  generating  stations  are  being 
operated  with  prime  movers,  boilers  etc.  underloaded 
because  the  100  percent  power-factor  generators  which 

TABLE  T— EFFECT  OF  POWER-FACTOR  ON  VOLTAGE 
DROP 


Percent  Power-Factor. 

100 

90 

80 

70 

Generators  *(oldcr  design) 
Transformers 
Transmission  line 

S,o 

1.2 

7-9 

4-1 
130 

1 
25.0 

4-9 
14.2 

1        - 
5-5 

15-2 

ihey  drive  limit  the  amount  of  power  that  can  be  gen- 
erated without  endangering  the  generator  windings. 
This  condition  some  times  makes  it  necessary  to  oper- 
ate three  units,  where  two  might  be  sufficient  to  carry 
the  load  at  unity  power-factor.  The  shutting  down  of 
r  unit  would  result  in  a  considerable  saving  in  steam 
consumption.  A  recent  case  came  up  of  a  transmis- 
sion line  30  miles  long,  fed  at  each  end  by  a  small  gen- 
erating station.  On  account  of  heavy  line  drop  it  was 
necessary  to  operate  both  stations  to  furnish  the  com- 
paratively light  night  load.  Investigation  developed 
that  by  installing  a  synchronous  condenser  at  one  of 
these  tenninal  stations  for  reducing  the  voltage  drop 
in  the  line,  one  generating  station  could  be  shut  down 
during  the  night,  thereby  resulting  in  a  very  large 
annual  saving  in  coal  and  labor  bills. 

A  station  may  have  some  generating  units  de- 
signed for  100  percent  power-factor  and  other  units 
designed  for  80  percent  power- factor;  or  again,  where 
two  generating  stations  feed  into  the  same  transmission 
system,  one  may  have  100  percent  power-factor  gen- 
erating units  and  the  other  80  percent  power-factor 


*The  present-day  design  of  maximum  rated  generators  with 
a  short-circuit  ratio  of  about  unity  will  barely  circulate  full- 
load  current  with  normal  no-load  excitation.  Under  such  con- 
ditions the  terminal  voltage  would  be  practically  zero  regardless 
of  the  oower-factor. 


136 


SVXCHRONOCS  MOTORS  AND  CONDENSERS    fOR    POWER-FACTOR    CORRECTION 


generating  units.  In  such  cases,  the  field  strength  of 
the  generators  may  be  so  adjusted  as  to  cause  the  80 
percent  power-factor  units  to  take  all  the  lagging  cur- 
rent, thus  permitting  the  100  percent  power-factor 
units  to  be  loaded  to  their  full  kw  rating. 

BEHAVIOR  OF  A.   C.   GENERATORS  WHEN   CHARGING  A 
TRANSMISSION    LINE* 

It  has  been  shown  above  how  leading  armature 
current,  by  increasing  the  field  strength,  causes  an  in- 
crease in  the  voltage  induced  in  the  armature  of  an 
alternator  and  consequently  an  increase  in  its  terminal 
voltage.  It  was  also  shown  that  the  terminal  voltage 
is  further  increased  as  result  of  the  voltage  due  to 
self  induction  adding  vectorially  to  the  voltage  induced 
in  the  armature. 

If  an  alternator  with  its  fields  open  is  switched 
onto  a  dead  transmission  line  having  certain  electrical 
characteristics,  it  will  become  self  exciting,  provided 
there  is  sufficient  residual  magnetism  present  to  start 
the  phenomenon.  In  such  case,  the  residual  magnet- 
ism in  the  fields  of  the  generator  will  cause  a  low  volt- 
age to  be  generated  which  will  cause  a  leading  line 
charging  current  to  flow  through  the  armature.  This 
leading  current  will  increase  the  field  flux  which  in 
turn  will  increase  the  voltage,  causing  still  more  charg- 
ing current  to  flow,  which  in  turn  will  still  further  in- 
crease the  line  voltage.  This  building  up  will  continue 
until  stopped  by  saturation  of  the  generator  fields. 
This  is  the  point  of  stable  operation.  Whether  or  not 
a  particular  generator  becomes  self  exciting  when 
placed  upon  a  dead  transmission  line  depends  upon  the 
relative  slope  of  the  generator  and  line  characteristics. 

In  Fig.  66  are  shown  two  curves  for  a  single 
45  000  kv-a,  1 1  000  volt  generator,  the  charging  current 
of  the  transmission  line  being  plotted  against  genera- 
tor terminal  voltage.  One  curve  corresponds  to  zero 
excitation,  the  other  curve  to  26.6  percent  of  normal 
excitation.  A  similar  pair  of  curves  correspond  to 
two  duplicate  generators  in  parallel**.  The  straight 
line  representing  the  volt-ampere  characteristics  of  the 
transmission  line  fed  by  these  generators  corresponds 
to  a  220  kv,  60  cycle,  three-phase  transmission  cir- 
cuit, 225  miles  long,  requiring  69  000  kv-a  to  charge  it 
with  the  line  open  at  the  receiving  end. 

The  volt-ampere  charging  characteristic  of  a 
transmission  line  is  a  straight  line,  that  is,  the  charg- 
ing current  is  directly  proportional  to  the  line  voltage. 
On  the  other  hand  the  exciting  volt-ampere  character- 
istic for  the  armature  has  the  general  slope  of  an 
ordinary  saturation  curve. 


*For  a  more  detailed  discussion  of  this  subject  see  the  fol- 
lowing articles : — "Characteristics  of  Alternators  when  Excited 
by  Armature  Currents"  by  F.  T.  Hague,  in  the  Journal  for  Aug. 
1915 ;  "The  Behavior  of  Alternators  with  Zero  Power-Factor 
Leading  Current"  by  F.  D.  Newbury,  in  the  Journal  for  Sept. 
1918;  "The  Behavior  of  A.  C.  Generators  when  Charging  a 
Transmission  Line"  by  W.  O.  Morris,  in  the  General  Electric 
Revinv  for  Feb.  1920. 

**It  is  assumed  that  with  the  assumed  field  current  such 
generators  can  be  synchronized  p.n<\  held  together  during  the 
process  of  charging  the  line. 


If  the  alternator  characteristic  lie  above  the  line 
characteristic  at  a  point  corresponding  to  a  certain 
charging  current  the  leading  charging  current  will 
cause  a  higher  armature  terminal  voltage  than  is  re- 
quired to  produce  that  current  on  the  line.  As  a  re- 
sult the  current  and  voltage  will  continue  to  rise  until, 
on  account  of  saturation,  the  alternator  characteristic 
falls  until  it  crosses  the  line  characteristic.  At  this 
point  the  voltage  of  the  generator  and  that  of  the  line 
are  the  same  for  the  corresponding  current.  If  on  the 
other  hand  the  alternator  characteristic  falls  below  the 
line  characteristic  the  alternator  will  not  build  up  with- 
out permanent  excitation. 

As  stated  previously,  whether  or  not  a  generator 
becomes  self-exciting  when  connected  to  a  dead  trans- 
mission line  depends  upon  the  relative  slopes  of  gener- 
ator and  transmission  line  characteristics.  The  rela- 
tive slopes  of  these  curves  depend  upon: — 

a — The  magnitude  of  the  line  charging  current. 

h — The  rating  of  the  generators  compared  to  the  full  voltage 
charging  kv-a  of  the  line. 

c — The  armature  reaction.  High  armature  reaction,  (that  is 
low  short-circuit  ratio)  favors  self-excitation  of  the  gener- 
ators. 

d — The  armature  reactance.  High  armature  reactance  also 
favors  self-excitation  of  the  generators. 

Methods  of  Exciting  Transmission  Lines — If  the 

relative   characteristics  of  an  alternator  and  line  are 

such  as  to  cause  the  alternator  to  he  self-exciting,  this 

condition  may  be  overcome  by  employing  two  or  more 


1000  2000  3000  4000  500C 

CHARGING  CURRENT  OF  LINE 

(AMPERES  PER  GENERATOR  TERMINAL) 


FIG.  66 — VOLT  AMPERE  CHARACTERISTICS  OF  ONE  45  OOO  KV-A,   1 1  COO 

VOLT  generator;  TWO  DUPLICATE  45  COO  KV-A  GENERATORS;   AND  A 

THREE-PHASE,  SINGLE-CIRCUIT,  220  KV  TRANSMISSION  LINE 

alternators  (provided  they  are  available  for  this  pur- 
pose) to  charge  the  transmission  line.  The  combined 
characteristics  of  two  or  more  alternators  may  be  such 
as  to  fall  under  the  line  characteristic,  in  which  case 
the  alternator  will  not  be  self-exciting.  In  such  case, 
the  alternators  could  be  brought  up  to  normal  speed, 
and  given  sufficient  field  charge  to  enable  them  to  be 


SYNCHRONOUS  MOTORS  AND  CONDENSERS    FOR    POWER-FACTOR    CORRECTION 


137 


TABLE  U— INSTALLATIONS  OF  LARGE  PHASE   MODIFIERS    (1921) 
By  American  Manufactnrers 


synchronized  and  held  in  step,  after  which  they  could  Fig.  66,  and  there  were  sufficient  residual  magnetism 
be  connected  to  the  dead  transmission  line  and  their  to  start  the  phenomenon,  the  generator  voltage  would 
voltage  raised  to  normal.  rise  to  approximately  double  normal  value  before  the 

Generators  as  normally  designed  will  carry  ap-  point  of  staple  operation  is  reached.  If,  however, 
proximately  40  percent  of  their  rated  current  at  zero  two  generators  having  26.6  percent  of  normal  excita- 
leading  power-factor.  If  more  than  this  current  is  tion  were  paralleled  and  connected  to  this  circuit,  a 
demanded  of  them  they  are  likely  to  become  unstable  point  of  staple  operation  would  be  reached  at  a 
in  operation.  By  modifying  the  design  of  normal  terminal  voltage  of  approximately  15  500  volts, 
alternators  so  as  to  give  low  armature  reaction,  they  Actually  stable  operation  would  be  reached  at  a  some- 
may  be  made  to  carry  a  greater  percentage  of  leading  what  less  terminal  voltage  for  the  reason  that  the  line 
current.     If  the  special  design  is  such  that  with  zero     would  probably  not  be  open  at  the  receiving  end,  but 

would  probably  have  the 
lowering  transformers  con- 
nected to  it.  In  such  case 
the  magnetizing  current 
required  for  lowering  trans- 
formers would  lower  the 
receiving  end  voltage,  re- 
sulting in  less  line  charging 
current. 

In  either  case  the 
curves  of  Fig.  66  show  that 
either  more  than  two  gen- 
erators will  be  required  to 
charge  the  line  when  un- 
loaded, or  some  other  meth- 
od of  charging  must  be  re- 
sorted to.  Reactance  coils 
could  be  used  at  the  receiv- 
ing end  t )  furnish  lagging 
current  for  neutralizing  some 
of  the  line  charging  cur- 
voltage  field  excitation  when  carrying  half  the  line  rent,  but  there  might  be  difficulty  in  removing  these 
charging  kv-a,  the  armature  voltage  will  not  exceed  70  from  the  circuit  when  the  line  is  fully  charged  At 
percent  of  normal,  this  reduced  voltage  will  result  in  the  present  time  it  is  expected  that  the  problem  of 
a  line  charging  kv-a  of  half  of  normal  value.  Spe-  charging  long  transmission  lines  may  usually  be  solved 
cially  designed  alternators  usually  result  in  larger  and  by  starting  one  or  more  generators  with  sufficient  field 
more  costly  machines  and  the  gain  resulting  in  the  spe-  strength  to  permit  them  to  be  synchronized  and  held  in 
cial  design  is  usually  not  sufficient  to  warrant  the  extra  step.  One  or  more  phase  modifiers  with  under-excited 
^°^^-  ,  fields  may  then  be  connected  to  the  line  at  the  receiving 

If  a  single  generator  with  its  field  circuit  open  end  and  brought  up  to  normal  speed  with  the  genera- 
were  connected  to  a  dead  transmission  circuit  such  as  tors.  Such  a  method  of  solving  this  problem  has  been 
the  one  whose  volt-ampere  characteristics  are  shown  in     employed  by  the  Southern  California  Edison  Company. 


Kv-a 

a.p.M. 

Volts 

Cycles 

No.  of 
Units 

Pate  of 
Order 

NAMF.  AND  LOCATION 

80  000 

600 

6600 

50 

1 

1919 

So.  Csl.  Ed.  Co.,  Los  Angeles,  Cal. 

20  000 

600 

11  000 

60 

2 

1921 

Pacific  Gas  &  Elec. 

15  000 

875 

6600 

50 

1912 

Southern  Oal.  Ed.  Co.,  Los  Ang.,  Cal. 

15  000 

375 

6600 

50 

1 

1912 

Pacific  Lt.  &  Pr.  Co. 

112  500 

500 

22  000 

50 

2 

1918 

Andhra  Valley,  India 

7500 

400 

6600 

60 

2 

1913 

Utah  Pr.  &  Lt.  Co.,   Salt  Lake,  Utah 

7500 

400 

6600 

60 

2 

1916 

Canton  El.  Co.,   Canton,   Ohio 

7500 

600 

13  800 

60 

1 

1917 

Blackstone  Valley  Gas  &  Elec.  Co.,  Pawtucket,  R. 

I. 

7500 

600 

13  800 

60 

1 

1917 

New  England  Pr.  Co.,  Worcester,   Mass. 

7500 

720 

13  800 

60 

1 

1918 

New  England  Pr.  Co.,   Fitchburg,  Mass. 

7500 

800 

11  500 

40 

1 

1918 

Adirondack  El.  Pwr.  Corp.,  Watervliet,  New  York 

7500 

750 

U  000 

50 

1 

1919 

Energia   Electrica   de   Cataluna,    Barcelona,    Spain 

7500 

600 

11  000 

60 

1 

1920 

Duquesne  Light  Co. 

7500 

600 

1200 

60 

2 

1918 

J.   G.    White,    Engineers 

7500 

600 

11  000 

60 

1 

1918 

Puquesne  Light  Co. 

7500 

600 

11  000 

60 

1 

1916 

Duquesne  Light  Co. 

7500 

600 

11  000 

60 

2 

1917 

Puquesne  Light  Co. 

6500 

750 

2200 

50 

1 

1917 

Shanghai  Municipal  Council,   Shanghai,  China 

6000 

500 

16  500 

50 

1 

1914 

So.  Cal.  Ed.  Co.,  Los  Angeles,  Cal. 

5000 

600 

7200 

60 

1 

1916 

Pac.  Pwr.  &  Lt.,  Kennewick,  Wash. 

5000 

500 

6600 

50 

2 

1915 

Tata  Hydro  El.  Pr.  &  S.  Co.,   India 

6000 

750 

6600 

50 

3 

1917 

Ebro  Irrigation  &  Pr.  Co.,  Barcelona,   Spain 

5000 

750 

11  500 

50 

1 

1919 

Societa    Lombarda   Distribuziona   Energia  Elettrica 

Italy 

6000 

600 

2300 

60 

1 

1918 

TurnbuU  Steel  Co.,  Warren,   Ohio 

5000 

720 

2300/ 
4000 

60 

1921 

Public  Service  of  N.  111. 

5000 

720 

11000 

60 

1 

1921 

Takata  &  Co.,  Japan. 

5000 

600 

13  200 

60 

1 

1919 

Conn.  Lt.  &  Pr.  Co. 

CHAPTER  XV 
PHASE  MODIFIERS  FOR  VOLTAGE  CONTROL 


WITH  alternating-current  transmission  there  is 
a  voltage  drop  resulting  from  the  resistance 
of  the  conductors,  which  is  in  phase  with  the 
current.  In  addition  there  is  a  reactance  voltage  drop ; 
that  is  a  voltage  of  self-induction  generated  within  the 
conductors  which  varies  with  and  is  proportional  to 
the  current,  and  may  add  to  or  decrease  the  line  volt- 
age. If  the  line  is  long,  the  frequency  high  or  the 
amount  of  power  transmitted  large,  this  induced  volt- 
age will  be  large,  influencing  greatly  the  line  drop. 
By  employment  of  phase  modifiers  the  phase  or  direc- 
tion of  this  induced  voltage  may  be  controlled  so  that 
it  will  be  exerted  in  a  direction  that  will  result  in  the 
desired  sending  end  voltage. 

A  certain  amount  of  self-induction  in  a  transmission 
circuit  is.  an  advantage,  allowing  the  voltage  at  the  re- 
ceiving end  to  be  held  constant  under  changes  in  load 
by  means  of  phase  modifiers.  It  may  even  be  made  to 
reduce  the  line  voltage  drop  to  zero,  so  that  the  voltage 
at  the  two  ends  of  the  line  is  the  same  for  all  loads. 
Self-induction  also  reduces  the  amount  of  current 
which  can  flow  in  case  of  short-circuits,  thus  tending  to 
reduce  mechanical  strains  on  the  generator  and  trans- 
former windings,  and  making  it  easier  for  circuit 
breaking  devices  to  function  successfully.  On  the 
ether  hand,  high  self-induction  reduces  the  amount  of 
power  which  may  be  transmitted  over  a  line  and  may, 
in  case  of  lines  of  extreme  length,  make  it  necessary 
to  adopt  a  lower  frequency.  It  also  increases  the  ca- 
pacity of  phase  modifiers  necessary  for  voltage  con- 
trol. High  reactance  also  increases  the  surge  over- 
voltage  that  a  given  disturbance  will  set  up  in  the  sys- 
tem. 

On  the  long  lines,  the  effect  of  the  distributed 
leading  charging  current  flowing  back  through  the  line 
inductance  is  to  cause,  at  light  loads,  a  rise  in  voltage 
from  generating  to  receiving  end.  At  heavy  loads,  the 
lagging  component  in  the  load  is  usually  sufficient  to 
reverse  the  low-load  condition;  so  that  a  drop  in  volt- 
age occurs  from  generating  to  receiving  end.  The 
charging  current  of  the  line  is,  to  a  considerable  extent, 
an  advantage;  for  it  partially  neutralizes  the  lagging 
component  in  the  load,  thus  raising  the  power-factor 
of  the  system  and  reducing  the  capacity  of  synchron- 
ous condensers  necessary  for  voltage  control. 

The  voltage  at  the  receiving  end  of  the  line  should 
be  held  constant  under  all  loads.  To  partially  meet 
this  condition,  the  voltage  of  the  generators  could  be 
varied  to  a  small  extent.  On  the  longer  lines,  how- 
ever, the  voltage  range  required  of  the  generators 
would    be    too    great    to    permit    regulation    in    this 


manner.  In  such  cases,  phase  modifiers  operating  in 
parallel  with  the  load  are  employed.  The  function  of 
phase  modifiers  is  to  rotate  the  phase  of  the  current  at 
the  receiving  end  of  the  line  so  that  the  self-induced 
voltage  of  the  line  (always  displaced  90  degrees  from 
the  current)  swings  around  in  the  direction  which  will 
result  in  the  desired  line  drop.  In  some  cases  a  phase 
modifier  is  employed  which  has  sufficient  capacity  not 
only  to  neutralize  the  lagging  component  at  full  load, 
but,  in  addition,  to  draw  sufficient  leading  current  from 
the  circuit  to  compensate  entirely  for  the  ohmic  and  re- 
actance voltage  drops  of  the  circuit.  In  this  case,  the 
voltage  at  the  two  ends  of  the  line  may  be  held  the 
same  for  all  loads.  This  is  usually  accomplished  by 
employing  an  automatic  voltage  regulator  which  oper- 
ates on  the  exciter  fields  of  the  phase  modifier.  The 
voltage  regulator  may,  if  desired,  be  arranged  to  com- 
pound the  substation  bus  voltage  with  increasing  load. 

CHECKING  THE  WORK 

A  most  desirable  method  of  determining  line  per- 
formance is  by  means  of  a  drawing  board  and  an  en- 
gineer's scale.  A  vector  diagram  of  the  circuit  under 
investigation,  with  all  quantities  drawn  to  scale,  greatly 
simplifies  the  problem.  Each  quantity  is  thus  repre- 
sented in  its  true  relative  proportion,  so  that  the  re- 
sult of  a  change  in  magnitude  of  any  of  the  quanti- 
ties may  readily  be  visualized.  Graphical  solutions 
are  more  readily  performed,  and  with  less  likelihood 
of  serious  error  than  are  mathematical  solutions.  The 
accuracy  attainable  when  vector  diagrams  are  drawn 
20  to  25  inches  long  and  accurate  triangles,  T  squares, 
straight  edges  and  protractors  are  employed  is  well 
within  practical  requirements.  Even  the  so-termed 
"complete  solution"  may  be  performed,  graphically 
with  ease  and  accuracy.  A  very  desirable  virtue  of 
the  graphical  solution  which  follows  is  that  it  exactly 
parallels  the  fundamental,  mathematical  solution.  For 
this  reason  this  graphical  solution  is  most  helpful  even 
when  the  fundamental  mathematical  solution  is  used, 
for  it  furnishes  a  simple  check  against  serious  errors. 
The  result  may  be  checked  graphically  after  each  in- 
dividual mathematical  operation  by  drawing  a  vector 
in  the  diagram  paralleling  the  mathematical  operation. 
Thus,  any  serious  error  in  the  mathematical  solution 
may  be  detected  as  soon  as  made.* 


*A  method  of  cfiecking  arithmetical  operations  which 
requires  little  time  and  is  an  almost  sure  preventative  of  errors 
is  that  known  as  "casting  out  the  nines."  This  method  is  given 
in  most  older  arithmetics  but  has  been  dropped  from  many  of 
the  modern  ones.  A  complete  discussion  is  given  in  Robinson's 
"New  Practical  Arithmetic"  published  by  The  American  Book 
Company. 


PHASE  MODIFIERS  I'OR  VOLTAGE  CONTROL 


139 


When  converting  a  complex  quantity  mathematic- 
ally from  polar  to  rectangular  co-ordinates,  or  vice 
versa,  the  results  may  readily  be  checked  by  tracing 
the  complex  quantity  on  cross-section  paper  and 
measuring  the  ordinates  and  polar  angle,  or  for  ap- 
proximate work  the  conversion  may  be  made  graphic- 
ally to  a  large  scale.  For  instance,  in  using  hyperbolic 
functions,  polar  values  will  be  required  for  obtaining 
powers  and  roots  of  the  complex  quantity.  For 
approximate  work  much  time  will  be  saved  by  ob- 
taining the  polar  values  graphically. 

In  the  graphical  solution  of  line  performance  it 
will  usually  be  desirable  to  check  the  line  loss  by  a 
mathematical  solution  in  cases  which  require  exact  loss 
values.  Since  the  line  loss  may  be  five  percent  or  less 
of  the  energy  transmitted,  a  small  error  in  the  overall 
results  might  correspond  to  a  large  error  in  the  value 
of  the  line  loss. 

EFFECT  OF  TRANSFORMERS   IN   THE   CIRCUIT 

Usually  long  transmission  circuits  have  trans- 
formers installed  at  both  ends  of  the  circuit  and  one 
or  more  phase  modifiers  in  parallel  with  the  load. 
Such  a  transmission  circuit  must  transmit  the  power 
loss  of  the  phase  modifiers  and  of  the  receiver  trans- 
formers. In  addition  to  this  power  loss,  a  lagging  re- 
active current  is  required  to  magnetize  the  transformer 
iron.  A  complete  solution  of  such  a  composite  cir- 
cuit (generator  to  load)  requires  that  the  losses  of  the 
phase  modifiers  and  transformers  be  added  vectorially 
to  the  load  at  the  point  where  they  occur  so  that  their 
complete  effect  may  be  included  in  the  calculation  of 
the  performance  of  the  circuit.  A  complete  solution 
also  requires  that  three  separate  solutions  be  made  for 
such  a  circuit.*  First  with  the  known  or  assumed  con- 
ditions at  the  load  side  of  the  lowering  transformers 
the  corresponding  electrical  conditions  at  the  high  volt- 
age side  of  the  transformers  is  determined  by  the  usual 
short  line  impedance  methods.  With  the  electrical 
conditions  at  the  receiving  end  of  the  high-tension  line 
thus  determined,  the  electrical  conditions  at  the  send- 
ing end  of  the  line  are  determined  by  one  of  the  vari- 
ous methods  which  take  into  account  the  distributed 
quantities  of  the  circuit.  With  the  electrical  condition 
at  the  sending  end  thus  determined  the  electrical  con- 
ditions at  the  generating  side  of  the  raising  transform- 
ers are  determined.  The  above  complete  method  of 
procedure,  is  tedious  if  carried  out  mathematically, 
but  if  carried  out  graphically  is  comparatively  simple. 

It  is  the  general  practice  to  neglect  the  effect  of 
condenser  and  lowering  transformer  loss  in  traveling 
over  the  line,  but  to  add  this  loss  to  the  loss  in  the 
high-tension  line  after  the  performance  has  been  calcu- 
lated. If  the  loss  in  condensers  and  lowering  trans- 
formers is  five  percent  of  the  power  transmitted  the 


*A  method  for  calculating  a  transmission  line  with  trans- 
formers at  each  end  in  one  solution  is  given  in  the  articles  by 
Messers.  Evans  and  Sels  in  the  Journal  for  July,  August,  Sept- 
ember, ct  scq.  1921. 


error  in  the  calculated  results  would  probably  be  less 
than  0.5  percent,  a  rather  small  amount. 

In  order  to  simplify  calculations,  it  is  the  general 
practice  to  consider  the  lumped  transformer  impedance 
as  though  it  were  distributed  line  impedance  by  add- 
ing it  to  the  linear  constants  of  the  line  and  then  pro- 
ceeding with  the  calculations  as  though  there  were  no 
transformers  in  the  circuit.  This  simplifies  the  solu- 
tion but  at  the  expense  of  accuracy,  particularly  if  the 
line  is  very  long,  the  frequency  high  or  the  ratio  trans- 
former to  line  impedance  high.  This  simplified  solu- 
tion introduces  maximum  errors  of  less  than  two  per- 
cent in  the  results  for  a  225  mile,  60-cycle  line. 

It  has  been  quite  general  practice  to  disregard  the 
effect  of  the  magnetizing  current  consumed  by  trans- 
formers. The  magnetizing  current  required  to  excite 
transformers  containing  the  older  transformer  iron  was 
about  two  percent  and  therefore  its  effect  could 
generally  be  ignored.  Later  designs  of  transformers 
employ  silicon  steel,  and  their  exciting  current  varies 
from  about  20  percent  for  the  smaller  of  distribu- 
tion type  transformers,  to  about  12  percent  on  trans- 
formers of  100  kv-a  capacity  and  about  five  percent  for 
the  very  largest  capacity  transformers.  The  average 
magnetizing  current  for  power  transformers  is  between 
six  and  eight  percent.  This  magnetizing  current  is  im- 
portant for  the  reason  that  it  is  practically  in  opposition 
to  the  current  of  over-excited  phase  modifiers  used  to 
vary  the  power-factor.  If  in  a  line  having  100  000 
kv-a  transformer  capacity  at  the  receiving  end,  the 
magnetizing  current  is  five  percent,  there  will  be  a 
5000  kv-a  lagging  component.  If  the  capacity  of 
phase  modifiers  required  to  maintain  the  proper  volt- 
age drop  under  this  load  is  50000  kv-a  the  lagging 
magnetizing  component  of  5000  kv-a  will  subtract  this 
amount  from  the  effective  rating  of  the  phase  modi- 
fiers, with  a  resulting  error  of  ten  percent  in  the  ca- 
pacity of  the  phase  modifiers  required. 

In  the  diagrams  and  calculations  which  follow,  the 
transformer  leakage,  consisting  of  an  in-phase  com- 
ponent of  current  (iron  loss)  and  a  reactive  lagging 
component  of  current  (magnetizing  current),  is  con- 
sidered as  taking  place  at  the  low-tension  side  of  the 
transformers.  A  more  nearly  correct  location  would 
be  to  consider  the  leak  as  at  the  middle  of  the  trans- 
former, that  is,  to  place  half  the  transformer  imped- 
ance on  each  side  of  the  leak.  To  solve  such  a  solu- 
tion it  would  be  necessary  to  solve  two  complete  im- 
pedance diagrams  for  the  transformers  at  each  end  of 
the  circuit.  The  gain  in  accuracy  of  results  would 
not,  for  power  transmission  lines,  warrant  the  in- 
creased arithmetical  work  and  complication  necessary. 

In  the  case  of  lowering  transformers,  it  would 
seem  that  the  magnetizing  current  would  be  supplied 
principally  from  synchronous  machines  connected  to 
the  load.  If  phase  modifiers  are  located  near  the 
lowering  transformers,  the  transformers  would  prob- 
ably  draw   most   of    their   magnetizing   current    from 


140 


PHASE  MODIFIERS  FOR  VOLTAGE  CONTROL 


them  rather  than  from  the  generators  at  the  distant 
end  of  the  Hne.  Partly  for  this  reason,  but  more 
particularly  for  simplicity,  the  leak  of  the  lowering 
transformers  will  be  considered  as  taking  place  at  the 
lead  side  of  the  transformers.     On  this  basis  we  first 


current  also  from  the  low  side;  that  iS  from  the  gen- 
erators. Both  the  complete  and  the  approximate 
methods  of  solving  long  line  problems  which  follow, 
include  the  effect  of  not  only  the  magnetizing  current 
consumed  by  the  transformers,  but  also  the  losses  in 


TABLE  V— COMPARISON  OF  RESULTS  AS  OBTAINED  BY  FIVE  DIFFERENT  METHODS  OF  CALCULATIONS 


76,000  KW  (88,236  KVA  AT  85%  PF)  3  PHASE  60  CYCLES     RECEIVER  VOLTAGE  HELD  CONSTANT  AT  220  KV,    60.000  KV-A  CONDENSER  AT  RECEIVING  END 
LENGTH  OF  TRANSMISSION  226  MILES      ALL  TABULATED  VALUES  REFERRED  TO  NEUTRAL 

AREA 

IN 

CIRCULAR 

MILS 

a 

RECEIVING  END  TO  NEUTRAL 

8E1IDINQ  END  TO  NEUTRAL 

LOSSES  IN  KW  TO  NEUTRAL                                                 1 

LOW  TENSION  SIDE  OF 
TRANSFORMERS. 

HIGH  TENSION  SIDE  OF 
TRANSFORMERS 

HIGH  TENStON  SIDE  OF 
TRANSFORMERS 

LOW   TENSION  SIDE  OF 
TRANSFORMERS 

LOWERING 
TRANSFORMERS 

CONDENSER 

HIGH          1         RAISING 
TENSION  LINEJTRANSFORMERS 

TOTAL  LOSS 

VOLTS 

4MP8 

PF. 

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VOLTS 

AMPS 

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CURRENT 

PFs 

LEAD 

VOLTAGE 

CURRENT 

P*bEN 

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COPPER 

KW, 

LOSS 

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*A — Transformer  impedances  treated  as  lumped  at  the  ends  of  the  line.  This  is  the  most  nearly  accurate  of  the  five  methods. 
It  is  referred  to  in  the  text  as  the  complete  solution. 

B — This  assumes  the  impedance  of  the  lowering  transformers  as  line  impedance.  It  takes  no  account  of  the  leakage  of  the 
lowering  transformers. 

C — This  assumes  the  impedance  of  both  lowering  and  raising  transformers  as  line  impedance — It  takes  no  account  of  the  leak- 
age of  the  lowering  and  raising  transformers. 

D — This  is  the  same  as  B  except  that  the  leakage  of  the  lowering  transformers  has  been  added  to  the  load — It  is  referred  to 
in  the  text  as  the  approximate  solution. 

E — This  is  the  same  as  C  except  that  the  leakage  of  the  lowering  transformers  has  been  added  to  the  load. 


have  a  load  current  expressed  in  rectangular  co- 
ordinates with  the  load  voltage  as  a  temporary  vector 
of  reference.  To  this  we  add  algebraically  a  phase 
modifier  current  (loss  -|-  j  or  leading)  and  to  this  we 
add  the  transformer  leakage  (loss  —  j  or  lagging).  In 
other  words,  these  three  components  of  current  at  the 
leceiving  end  of  the  line  add  up  algebraically  upon  a 


transformers  and  phase  modifiers  flowing  over  the  line. 
For  the  purpose  of  determining  the  magnitude  of 
errors  in  the  calculated  results  corresponding  to  sim- 
plified methods  of  calculation  where  transformers  are 
required  at  both  ends  of  the  line,  the  calculations  shown 
in  Table  V  were  made.  Five  methods  of  calculations 
were  made  for  each  of  four  sizes  of  cable.     A  con- 


TABLE  W-PERCENTAGE  ERRORS  IN  RESULTS,  AS  DETERMINED  BY  VARIOUS  METHODS  OF  CALCULATION. 
These  methods  do  not  take  complete  account  of  the  effects  of  the  transformers  in  the  circuit 


Method 

At  Generator 
Percent    Error 

At  Sending^  End 
Percent    Error 

Line  Loss 

Percent 

Error 

Transformer  Account 

Egen 

Igen 

PPe... 

E,  !   I,    '  pFk 

A 

0 

0 

0 

0,0       0 

1 

„          ;    Complete  method — Assumed  for  comparison  as  resulting  in  100 
percent  values. 

B 

-3.7 

+3.9-0.42 

,  „  „-          Leak  of  lowering  transformers  ignored.     Impedance  of  lowering 
"•""•"^               transformers  assumed  as  line  impedance. 

C 

-1.8 

+2.8 

-2.35 

... 

,f.  ._          Leaks  of  raising  and  lowering  transformers  ignored.     Impedance 
"■"                      of  raising  and  lowering  transformer  assumed  as  line  impedance. 

D 

-1.6 

+0.4 

+0.55 

,  r,  „.       !   Same  as  B  except  that  the  transformer  leak  has  been  added  to 
+0-°"^              the  load. 

E 

+0.5 

-0.7 

-1.62 

1 

Same  as  C  except   that  the  transformer  leak  has  been   added 
-^•^^              to  the  load. 

common  vector  of  reference,  thus  making  it  very  easy 
to  obtain  the  resulting  load  at  the  receiving  end  of  the 
line. 

The  transformers  at  the  sending  end  of  the  line 


stant  load,  load  voltage  and  condenser  capacity  were  as- 
sumed for  all  calculations  and  the  results  of  these  calcu- 
lations are  tabulated  in  Table  W.  Thus  method  D  which 
does  not  take  anv  account  of  the  lowering  transformer 


have   been   considered   as   receiving   their   magnetizing     magnetizing  current  and  assumes  the  transformer  im- 


PIlASli  MODfl-IKRS  FOR  VOLTAGE  CONTROL 


MI 


pedance  as  line  impedance,  gives  the  sending  ena  volt- 
age too  low  by  3.7  percent  and  the  current  too  high  by 
3.9  percent. 

Table  X  contains  approximate  data  upon  trans- 
formers of  various  capacities  25  and  60  cycles.  Since 
such  data  will  vary  greatly  for  different  voltages  it 
must  be  considered  as  very  approximate  but  may  be 
found  useful  in  the  absence  of  specific  data  for  the 
problem  at  hand. 

Fig.  67  shows  complete  current  and  voltage  dia- 
grams for  both  short  and  long  lines.  The  diagram 
illustrating  short  lines  is  based  upon  the  current  hav- 
ing the  same  value  and  direction  at  all  points  of  the 
circuit.  On  this  basis  the  IR  drops  of  the  line  and  of 
the  raising  and  lowering  transformers  will  be  in  the 
same  direction.  Likewise  their  individual  IX  drops 
will  also  be  in  the  same  direction.  It  is  evident, 
therefore,  that,  for  short  lines  where  the  capacitance 


voltage  circuit  in  order  to  combine  properly  with  the 
linear  constants  of  the  line.  Although  all  calculations 
are  made  in  terms  of  the  high-voltage  circuit  the  re- 
sults may,  if  desired,  be  converted  to  terms  of  the  low 
voltage  circuit,  by  applying  the  ratio  of  transformation. 
The  transformer  impedance  to  neutral  is  one- 
third  the  equivalent  single-phase  value.  The  reason 
for  this  is  that  the  I^R  and  I^X  for  one  phase  is 
identical  whether  to  neutral  or  between  phases.  Since 
the  current  between  phases  is  equal  to  the  current  to 
neutral  divided  by  VS.  the  square  of  the  phase  current 
would  be  one-third  the  square  of  the  current  to  neu- 
tral ;  therefore,  R  and  X  to  neutral  will  be  one-third  the 
phase  values.  Another  way  of  looking  at  this  is  that 
the  resistance  and  reactance  ohms  vary  with  the  square 
of  the  voltage,  and  since  the  phase  voltage  is  V^  times 
the  voltage  to  neutral,  the  phase  resistance  and  phase 
reactance  would  be   three  times  that  to  neutral.     In 


TABLE  X— APPROXIMATION  OF  RESIS'^ANCE  AND  REACTANCE  VOLTS,  OF  IRON  AND  COPPER  LOSSES 
AND  OF  MAGNETIZING  CURRENT  FOR  TRANSFORMERS  OF  VARIOUS  CAPACITIES 


Capacity 

of 

Transformer 

KV-A 

60  CYCLES  PER  SEC.O.VD 

25  CYCLES  PER  SECO.ND 

Percen : 
'Resistance 

Percent 
■Reactance 

Percent    Loss 

Percen  t 

Magnetizing 

Current 

Percent 
Resbtance 

Percent 
Re  ictance 

Percent  Loss 

Pirienl 

Muunclizing 

Current 

Iron           Copper 

Iron       j     Copper 

200 
30J 
500 

1.5 

1.3 
1.2 

5.5  1.4 

5.6  1.3 
6.0                      1.2 

1 

l.S 
1.3 
1.2 

10 
9 

8 

2.6 
2.15 
1.85 

4.0                     1.1                   2.6 
4.0                     1.0                   2.1s 
i.l                     1.0         ,          1.85 

i 

10 
10 
9 

780 
1000 
1500 

1.1 
1.1 
0.9 

6.3 
6.5 
7.0 

1.0 
0.9 
0.8 

1.1 
1.1 
0.9 

8 
7 
6 

l.«S 
1.55 
1.4 

i2                    0.9                   1.65 
t.O                     0.8                   1.55 
6.2                     0.8                   1.4 

9 
8 
8 

2000 
3000 
MlflO 

0.8 

0.75 

0.65 

7.0 
7.0 
7.0 

0.7 
0.7 
0.6 

0.8 

0.75 

0.65 

6 
6 
6 

1.3 
1J2 
1.1 

6.4 
6.8 
7.2 

0.7                     1.3 
0.6                    ISt 
0.5                    1.1 

8 
7 
7 

7500 
10000 
15000 

0.6 
0.6 
0.55 

3.0 

8.0 
8.5 

0.6         ,         0.6 
0.5                 0.6 
0.5                  0.55 

5 
5 
5 

1.0 
1.0 
0.96 

7.8 
8.0 
8.0 

0.5                     1.0 
0.5                     1.0 
0.6                    0.95 

7 
6 
6 

25000 
3.5000 
60000 

0.5 
0.5 
0.5 

9.0 
9.5 
10.0 

0.6         1         0.5 
0.6         i         0.5 
0.6                  0.5 

S 
5 
5 

0.9 
0.9 
0.9 

8.0            1         O.t                   0.9 
9.0                      0.6                    0.9 
9.0                      0.6                    0.9 

t 
6 
6 

•The  actual  ohms  resistance  and  ohms  reactance  will  vary  as  the  square  of  the  voltage.    The  values  in  above  table-  must  be  considered  as  only  roughly 
approximate.    They  will  vary  materially  with  transformers  wound  for  different  voltages 


is  neglible,  the  transformer  impedance  may  be  added 
directly  to  the  line  impedance,  provided  the  electrical 
characteristics  on  the  high-tension  side  of  the  trans- 
formers are  not  required. 

As  the  line  becomes  longer,  the  current  changes  in 
both  amount  and  direction  from  point  to  point,  as  a  re- 
sult of  the  superimposed  distributed  charging  current 
of  the  line.  The  result  of  this  is  that  the  impedance 
triangles  of  the  line  and  of  lowering  and  raising  trans- 
formers change  in  both  size  and  relative  position;  so 
that  their  individual  impedances  can  no  longer  be 
added  together  and  considered  as  all  line  impedance, 
without  accepting  an  error  in  the  results  thus  obtained. 
The  complete  diagram  for  long  lines  shown  by  Fig.  67 
will  be  considered  later. 

TRANSFORMER   IMPEDANCE  TO    NEUTRAL* 

Transformer  constants  are   referred   to   the  high 


*The  writer  desires  to  expre.ss  his  appreciation  of  helpful 
assistance  and  useful  data  on  transformer  characteristics  receiv- 
ed from  Mr.  J.  F.  Peters. 


calculating  the  impedance  to  neutral,  the  results  will  be 
the  same  whether  star  or  delta  connection  is  used. 

Even  if  the  transformers  at  both  ends  of  the 
transmission  line  are  duplicates  their  impedance  will 
not  be  the  same  if  operated  on  different  taps  of  the 
windings  to  accommodate  different  voltages.  In  such 
cases,  their  impedances  will  vary  as  the  square  of  the 
voltages.  For  instance,  if  they  are  operated  at  220 
and  230  kv  at  the  receiving  and  sending  end  respec- 
tively, then  their  impedances  will  have  the  relation  of 

220^  ' 

5^=  o.oit;.     In  other  words,  if  the  resistance  and 

230^^ 

reactance  of  the  receiving  end  transformers  is  3.185 
and  39.82  ohms  respectively,  the  sending  end  trans- 
formers will  have  resistances  and  reactances  of  3.481 
and  43.52  ohms  respectively;  provided  transformer 
taps  corresponding  to  this  higher  voltage  are  used.  <« 
The  impedance  in  ohms  of  an  18  000  kv-a,  three- 
phase,  or  of  three  6000  kv-a  single-phase  transform- 
ers, connected  in  a  bank,  may  be  determined  as  fol- 


J42 


PHASE  MODIFIERS  FOR  VOLTAGE  CONTROL 


lows.  Assume  that  they  are  operated  at  104000  volts 
between  conductors  (60  046  to  neutral)  and  that  the 
resistance  voltage  is  1.04  percent  and  reactance  voltage 
is  4.80  percent. 

The  single-phase  values  are: — 
6  000  000 


>?.■ 


A-,= 


10^  000 
10 i 000  X  0.0104 

104000  X  0.04S 


=  sy.j  amperes 


57-7 


=  1S.7S  ohms  resistance 


=  S6.52  ohms  reactanee 


The  values  to  neutral  are,  as  stated  above,  one- 
third  of  the  above ;  but,  for  the  sake  of  uniformity  in 
determining  values  to  neutral,  should  preferably  be  de- 
termined as  follows: — 


6  000  000 
^        .    =  99.92  amperes  to  neutral 


^tn  = 


^Vtn 


60  046  X  o.oro4 

99.92 
60046  X  0.0480 

99-92 


=  6.25  ohms  resistance  to  neutral 


2S.S4  ohms  reactance  to  neutral 


If  two  or  more  banks  operate  in  parallel,  the  re- 
sulting impedance  Z,.  can  be  obtained  by  taking  the  re- 


to  the  same  kv-a  base.  For  instance,  if  a  6000  kv-a 
and  a  3000  kv-a  transformer  each  have  a  resistance  of 
1.04  percent  and  a  reactance  of  4.8  percent,  their  im- 
pedance is  4.91  percent.  Before  combining  the  imped- 
ances, that  of  the  3000  kv-a  unit  should  be  put  in  terms 
of  the  6000  kv-a,  and  the  resultant  would  be: — 
4.91  X  9.S2 


Zr  = 


■■  J .27 percent  at  6000  kva. 


4.91  -f  9.S2  ■ 
=  0.159  percent  resistance  volts  at  6000  kva. 
—  3 .'9 percent  reactance  volts  at  6000  kva. 

If  the  impedance  triangles  of  the  two  banks  to  be 
paralleled  are  considerably  different  (that  is  their  ratio 
of  resistance  to  reactance)  it  will  be  necessary  to  ex- 
press the  impedances  in  complex  form.  We  have  as- 
sumed above  that  the  triangles  are  proportional,  other- 
wise they  would  not  divide  the  load  evenly  at  all 
power- factors.  Solving  the  preceding  problem  for 
the  resultant  impedance  by  complex  notation,  we  get: 

{1.04^  J4^)  X  {2.o8-\-J9.6) 
^'  -  {1.04  -h  J4.8)  -I-  {2.0S  +  J9.6) 
^  -43-9'7  -^  j^9.968 
3.12  +  j/4.4 


COMPLETE  DIAGRAM  FOR 
SHORT   LINES 


COMPLETC  DIAGRAM  POB 
LOWG  LINES 


IMPEDANCe  OF 

RAISING 

TRANSFORMERS 


LEAKAGE  OF 

RAISING 

TRANSFORMERS 


IMPEDANCE  OF 

RAISING 

TRANSFORMERS 


IMPEDANCE  OF 

LOWERING 
TRANSFORMERS 


THE  BROKEN  LINES  REPRESENT  THE  VOLTAGE 
DIAGRAM  FOR  THE  APPROXIMATE  SOLUTION 
USUALLY  employed;  THAT  IS,  WHEN  THE  IMPEO- 
ANCE  OF  THE  LOWERING  TRANSFORMERS  IS  ADD- 
ED TO  AND  CONSIDERED  AS  DISTRIBUTED  LINE 
IMPEDANCE-A  SIMILAR  DIAGRAM  MAV  BE  DRAWN 
TO  INCLUDE  THE  IMPEDANCE  OF  BOTH  THE  LOW- 
ERING AND  THE  RAISING  TRANSFORMERS-  IN  THE 
CASE  ILLUSTRATED  '  A  220  K«  225  MILE  6Q  CYCLE 
CIRCUIT!  THE  APPROXIMATE  SOLUTION  GIVES  * 
VOLTAGE  AT  THE  SENDING  END  TOO  LOW  BY  t  93% 
(THE  DISTANCE  BETWEEN  S'-  S). 


FIG.    67 — VECTOR   DIAGRAMS    FOR    SHORT    AND   LONG   LINES 


ciprocal  of  the  sum  of  the  reciprocals  of  the  individual 
impedance.     Thus: — 

Z\  Zi 


Zr  = 


Z, -t-Za 


In  the  above  example  Zt  =  T  I.O^  -\-  4.8^  = 
4.91  percent  . 

To  parallel  two  banks  containing  transformers 
duplicates  of  the  above,  we  get,  by  the  above  rule,  the 
following  resultant  impedance: — 


Z,= 


4.91  X  4.91 


=  2.4^  percent 


4-9'  -H  4-9'  ' 

Which  is  just  half  the  impedance  of  a  single  bank, 
as  is  evident  without  applying  the  rule. 

Where  two  or  more  banks  are  to  be  operated  in 
parallel  consisting  of  transformers  not  duplicates,  then 
the  above  rule  must  be  applied  to  determine  the  re- 
sultant impedance.  If  the  impedances  are  expressed 
in  percent,  as  is  usual,  then  they  must  be  both  referred 


^  48. 25  \iS5''32'5S" 

'4.734  /77''46'29" 
"  3.27  /77''l6'29"  ohms 
=  0.69  +  j 3.19  ohms 

Which  checks  with  the  results  determined  above 
on  the  percentage  basis. 

THE  AUXILIARY   CONSTANTS 

The  graphical  construction  for  short  lines  repre- 
sented typically  by  the  Mershon  Chart  is  so  generally 
known  and  understood  that  a  similar  construction 
modified  to  take  into  accurate  account  the  distribution 
effect  of  long  lines  will  readily  be  followed.  Both  the 
short  and  the  long  line  diagrams  are  reproduced  in  Fig. 
68.  From  these  diagrams  it  will  be  seen  how  the 
three  auxiliary  constants  correct  or  modify  the  short 
line  diagram  adapting  it  to  long  line  problems.  The 
two  mathematical  and  three  graphical  methods  of  ob- 
taining the   auxiliary   constants   are   Indicated   a*   the 


PHASE  MODIFIERS  FOR  VOLTAGE  CONTROL 


143 


bottom  of  this  figure.  Since  the  auxiliary  constants 
are  functions  of  the  physical  properties  of  the  circuit 
and  of  the  frequency  only,  they  are  entirely  independ- 
ent of  the  voltage  or  the  current.     Having  determined 


VECTORS  BASED  UPON  ONe  VOLT  AND  ONE  AMPERe  AT    B5»    POWER  f  ACTOR  BEINO 
OeUVEREO  AT  THE  RECEIVING  END-THE  DIAGRAMS  CORRESPOND  TO  A  LONG  LINE 


E8-ER(a,*ia2)*|R(t>,*it'2) 
ls-lR(3fja2)*ER(c,*jC2) 


QIAORAM  rOR  SHORT  LINES 


REACTANCE  VOLTS 

OF  H  T   LINE 

IN  QUADRATURE 

V^iTH  CURRENT 


RESISTANCE  rfOLTS 

OFH.T  LINE 

IN  PHASE 

WITH  CURRENT 


DtAORAM  FOR  LONG  LINES 
I  ZERO  LOAD) 
-  CURRENT  AT  SENDlNG-END 


■   CONSTANT  (CI 


^^^m^^msm^^M^m^^ 


(VECTOR  OF  REFERENCE) 


CHARGING  CURRENT 
RESISTANCE  VOLTS 
OF  H,T  LINE-  R'-F 


CHARGING  CURRENT 
REACTANCE  VOLTS 
OF  H  T,  LINE  -  F  -  R 


■  VOLTAGE  AT  REOEIVtNQ-ENO  - 1  VOLT- 


DIAQRAM  FOR  LONG  LINES 
(LOAD   CONDITION) 


THE  DOTTED  LINES 

CORRESPOND  TO  A  CIRCUIT 

DEVOID  OF  CHARGING  CURRENT. 

THAT  18-A  SHORT  LINE 


HOW  THE  AUXILIARY  CONSTANTS  MAY  BE  OBTAINED 


=  COSHB   (BY  REAL  HYPERBOLIC  FUNCTIONS-SEE  CHART  XVI) 
=   y^^-gjg  <GRAPHICAL-SEE  KENNELLY  CORRECTING  FACTOR  CHARTS  XVIIIXIX-XX-XXI) 
,"  COSH  e  (GRAPHICAL-SEE  KENNELLY  S  CHART  ATLAS.  HARVARD  PRESS) 
-  COSH  e  (ALL  GRAPHICAL  FROM  WILKINSON  8  CHART  "A '-SEE  CHART  V) 


(B)-(^l*i''2)=  z['*  -6-+ ^725- ♦g^,*352:geo*ETC.l(BY  CONVERGENT  SERIES-SEE  CHARTXI> 

-  Iy  SINH  e  (BY  REAL  HYPERBOLIC  FUNCTIONS-SEE  CHART  XVI) 

=    Z  51!^  (GRAPHICAL-SEE  KENNELLY  S  CORRECTING  FACTOR  CHARTS  XVIII  ■  XIX  I 

-  |y  SINH  e  (GRAPHICAL-SEE  KENNELLYS  CHART  ATLAS,  HARVARD  PRESS) 

-  1^  SINH  e  (ALL  GRAPHICAL  FROM  WILKINSON  S  CHART    B'  -SEE  CHART  VI) 

/      \     /  \  r        YZ     Y^Z'      Y^Z^       Y*Z*  T 

(CJ-(*=l'i*^)-   ^  L"    6'*T20'  *  6:3So*362ll8(f  ^"'■^J'^^  CONVERGENT  SERIES-SEE  CHARTXI) 

SINH  9  ;BY  REAL  HYPERBOLIC  FUNCTIONS-SEE  CHART  XVI) 

51Mi  (GRAPHICAL -SEE  KENNELLY  S  CORRECTING  FACTOR  CHARTS  XVUI  ■  XIX  ) 
SINH  e  (GRAPHICAL -SEE  KENNELLY'3  CHART  ATLAS.  HARVARD  PRESS) 

SINH  S  (ALL  GRAPHICAL  FROM  WILKINSONS  CHART  '  C"-8EE  CHART  Vli ) 


-       Y 

WHERE  9  -  Vz7   1^ 

FIG.   68 — HOW    THE    AUXILIARY    CONSTANTS     MODIFY    SHORT    LINE 
DIAGRAMS   ADAPTING  THEM    TO  LONG  LINE   PROBLEMS 

by  any  of  the  five  methods  referred  to,  the  value  for 
the  auxiliary  constants  corresponding  to  a  given  cir- 
cuit, the  remainder  of  the  solution  for  any  receiving 
end  current  or  voltage  is  readily  performed  graphically. 


Constants  Oj  and  a^ — If  the  line  is  short  electric- 
ally the  charging  current,  and  consequently  its  effect 
upon  the  voltage  regulation  is  small.  In  such  a  case 
constant  a^  would  be  unity  and  constant  Oj  would  be 
zero,  and  the  line  impedance  triangle  would  be  attached 
to  the  end  of  the  vector  ER  representing  the  receiving 
end  voltage,  since  this  vector  also  represents  the  send- 
ing end  voltage  at  zero  load. 

If,  however,  the  circuit  contains  appreciable  ca- 
pacitance, the  e.m.f.  of  self-induction  resulting  from 
the  charging  current  will  result  in  a  lower  voltage  at 
zero  load  at  the  sending  end  than  at  the  receiving  end 
of  the  line.  Obviously,  the  load  impedance  triangle 
must  be  attached  to  the  end  of  the  vector  representing 
the  voltage  at  the  sending  end  of  the  circuit  at  zero 
load.  This  is  the  vector  ER'  of  the  long  line  diagrams 
of  Fig.  68.  In  such  a  circuit  the  effect  of  the  charg- 
ing current  is  sufificiently  great  to  cause  the  shifting  of 
the  point  R  for  a  short  line  to  the  position  R'  for  the 
long  line.  The  constants  Oj  and  a^  therefore,  deter- 
mine the  length  and  position  of  the  vector  representing 
the  sending  end  voltage  at  zero  load.  Actually  the 
constant  Oj  represents  the  volts  resistance  drop  due  to 
the  charging  current  for  each  volt  at  the  receiving  end 
of  the  circuit.  That  is,  the  line  FR'  equals  approxi- 
mately one-half  the  charging  current  times  the  resist- 
ance R,  taking  into  account,  of  course,  the  distributed 
nature  of  the  circuit.  For  a  short  line,  it  would  be 
sufficiently  accurate  to  assume  that  the  total  charging 
current  flows  through  one-half  the  resistance  of  the 
circuit.  To  make  this  clear,  it  will  be  shown  later 
that,  for  a  220  kv  problem,  the  resistance  per  conduc- 
tor is  R  =  34.65  ohms  and  the  auxiliary  constant  C^  = 
O.001211  mho.  Thus,  this  line  will  take  0.001211  am- 
pere charging  current,  at  zero  load,  for  each  volt  main- 
tained at  the  receiving  end,  and  since  FR'  ;=  approxi- 
R 


mately  I^e  X  —  we  have  FR'  or 


0.001211   X 


(A)-(ac*jaj)»   [i  •  y*-54-  ♦755"*  SOm*  "°n'°''°°''^^''°E'''''8E"'E8-8EE0t«RTXI)  34-05 


=  0.020980.     The  exact  value  of  a^  as  calculated 


by  hyperbolic  functions,  taking  into  account  the  dis- 
tributed nature  of  the  circuit  is  0.020234.  Since  the 
charging  current  is  in  leading  quadrature  with  the 
voltage  ER,  the  resistance  drop  FR'  due  to  the  charg- 
ing current  is  also  at  right  angles  to  ER. 

The  length  of  the  line  FR  or  (one-Oj),  represents 
the  voltage  consmed  by  the  charging  current  flowing 
through  the  inductance  of  the  circuit.  This  may  also 
be  expressed  with  small  error  if  the  circuit  is  not  of 

X 
great  electrical  length  as  I^c  X — •     The  reactance  per 

conductor    for    the    220   kv    problem    is    178.2    ohms. 


Therefore,  FR  =  0.001211   X 


178.2 


=  0.107900  and 


a,  =   I  —  0.107900  =  0.892100.     The  exact  value 
of  a,  as  calculated  rigorously,  is  0.893955. 

Constants   fc,   and    b^ — These   constants   represent 
respectively  the  resistance  and  the  reactance  in  ohms. 


144 


PHASE  MODIFIERS  FOR  VOLTAGE  CONTROL 


as  modified  by  the  distributed  nature  of  the  circuit. 
The  values  for  these  constants,  multiplied  by  the  cur- 
rent in  amperes  at  the  receiving  end  of  the  circuit,  give 
the  IR  and  IX  volts  drop  consumed  respectively  by  the 
resistance  and  the  reactance  of  the  circuit.  To  illus- 
trate this,  the  values  of  R  and  X  for  the  220  kv  prob- 
lem are  34.65  ohms  and  178.2  ohms  per  conductor. 
The  distributed  effect  of  the  circuit  modifies  these 
linear  values  of  R  and  X  so  that  their  effective  values 
are  b-^  =  j^./pS"  and  tj  =  172.094  ohms.  The  line 
impedance  triangle,  as  modified  to  take  into  exact  ac- 
count the  distributed  nature  of  the  circuit,  is  therefore 
smaller  than  it  would  be  if  the  circuit  were  without 
capacitance. 

Constants  c^  and  c^ — These  constants  represent 
respectively  the  conductance  and  susceptance  in  mhos 
as  modified  by  the  distributed  nature  of  the  circuit. 
The  values  for  these  constants,  multiplied  by  the  volts 
at  the  receiving  end  of  the  circuit,  give  the  current  con- 
sumed respectively  by  the  conductance  and  the  suscep- 
tance of  the  circuit.  To  illustrate,  the  linear  value  of  c^ 
for  the  220  kv  problem  is  0.001211  mho.  The  distribu- 
tion effect  of  the  circuit  modifies  this  linear  value  so 
that  its  effective  value  Cj  =  0.001168.  The  value  of  c^ 
is  so  small  that  its  effect  is  negligible  for  all  except  for 
long  circuits.  An  exception  to  this  statement  would 
be  that  if  the  line  loss  is  very  small  compared  to  the 
amount  of  power  transmitted  the  percent  error  in  the 
value  of  line  loss  may  be  considerably  increased  if  the 
effect  of  Cj  is  not  included  in  the  solution.  If  c^  is 
ignored,  Cj  will  represent  the  charging  current  at  zero 
load  per  volt  at  the  receiving  end.  Thus  Cj  multiplied 
by  the  receiving  end  voltage,  gives  the  charging  current 
at  zero  load  for  the  circuit.  For  the  220  kv  problem 
Cj  =  0.001168  and  this  multiplied  by  127020,  the  re- 


ceiving end  voltage  to  neutral,  gives  148.36  amperes 
charging  current  per  conductor. 

Referring  to  the  formulas  at  the  top  of  Fig.  68, 
•'^r  (Oi  -|-  /  02)  is  that  part  of  E^  which  would  have  to 
be  impressed  at  the  sending  end  if  /r  =  0,  or  the  line 
was  freed  at  the  receiving  end  with  E^  steadily  main- 
tained there.  It  may  be  called  "free"  component  of 
£s*.  Again  /,  {b^  -\-  j  b^)  is  that  other  part  of  £, 
which  would  have  to  be  impressed  at  the  sending  end, 
if  £r  ^  0,  or  the  line  was  short-circuited  at  the  receiv- 
ing end,  with  I^  steadily  maintained  there.  It  may  be 
called  the  "short"  component  of  E^. 

Similarly,  the  term  /,  (Oi  +  /  a^)  is  the  compon- 
ent of  /s  necessary  to  maintain  /,  at  the  receiving  end 
without  any  voltage  there  (£,  =  0)  ;  while  E^  (q  + 
y  Cj)  is  the  component  of  I^  necessary  to  maintain  E^ 
at  the  receiving  end  without  any  current  there  (/,  = 
0).  The  reason  that  c^  is  likely  to  be  negative  in  ordi- 
nary power  lines  is  because  the  complex  hyperbolic 
angle  of  any  good  power  transmission  line  has  a  large 
slope,  being  usually  near  88  degrees.  The  sinh  of  such 
an  angle,  within  the  range  of  line  lengths  and  sizes  of  6 
ordinarily  present,   is  also  near  90  degrees  in  slope. 

The  surge  impedance  Zo  =  -^ —  of  such  a  line  is  not 

far  from  being  reactanceless ;  but  it  usually  develops  a 

small  negative  or  condensive  slope.  This  means  that  the 

I 
surge  admittance   Fo  ^  ^   usually  develops  a   small 

positive  slope.  Consequently,  C  or  the  product  E^ 
(Tj  -}-  /  '^2)  usually  slightly  exceeds  90  degrees  in 
slope ;  or  c^  becomes  a  small  negative  rectilinear  com- 
ponent. 


*See  paper  by  Houston  and  Kennelly  on  "Resonance  in  A. 
C.  Lines"  in  Trans.  A.  I.  E.  E.  April.  1895 


CHAPTER  XVI 
A  TYPICAL  220  KV  PROBLEM 


TO  illustrate  the  method  of  determining  the  per- 
formance of  long  lines  requiring  phase  modifiers 
for  voltage  control,  the  following  220  kv  prob- 
lem will  be  considered,  which  is  typical  of  many  like- 
ly to  be  considered  in  the  near  future.  A  line  necessi- 
tating such  large  expediture  would  warrant 
a  thorough  investigation  before  determining  the  final 
design.  The  conclusions  are  given  only  for  the  pur- 
pose of  illustrating  the  procedure. 

The  Problem — It  is  assumed  that  300000  kw  at 
85  percent  lagging  power-factor  is  to  be  delivered  a 
distance  of  225  miles,  at  220  kv,  three-phase,  60  cycles. 
Two  lines  will  be  required,  so  that  in  case  one  is  under 
repair,  the  other  will  transmit  the  entire  300000  kw 
load.  Since  the  self -induced  voltage  would  be  exces- 
sive if  the  300000  kw  were  transmitted  in  emergency 
over  a  single-circuit  tower  line,  we  will  as- 
sume that  each  tower  line  will  support  two 
three-phase  circuits.  The  cost  of  two  three- 
phase  circuits  per  tower  line  will  not  be 
greatly  in  excess  of  a  single  circuit  tower 
line  employing  conductors  of  double  the 
cross-section.  On  this  basis  each  of  the  four 
three-phase  circuits  will  normally  transmit 
75  000  kw  and,  under  emergency  condition, 
each  of  the  two  circuits  on  one  tower  line 


In  Table  Y  is  shown  a  comparison  of  values  of  cap- 
italized losses  vs.  first  costs  of  conductors  for  four 
sizes  of  aluminum-steel  cables  considered  in  connection 
with  this  220  kw  problem**.  The  cost  of  power  losses 
is  based  upon  rates  of  0.3,  0.4  and  0.5  cents  per  kw 
hour,  an  average  load  corresponding  to  80  percent  of. 
the  full  load  loss  and  a  capitalization  of  these  losses 
at  15  percent.  The  cost  of  the  cables  is  based  upon 
29  cents  per  pound  for  the  complete  cable  (aluminum 
plus  the  steel).  All  tabulated  data  is  based  upon  four 
three-phase  circuits.  The  losses  include  those  in  the 
high  voltage  line  only.  If  the  capacity  of  transformers 
or  phase  modifiers  varies  materially  for  different  con- 
ductors, the  difference  in  their  losses  should  be  included. 
If  the  base  load  power  generated  in  such  a  large 
amount  by  water  power  costs  0.3  of  a  cent  per  kw-hr., 


will  transmit  150000  kw.     Such  a  transmis- 
sion is  illustrated  by  Fig.  69* 

Economic  Size  of  Conductors — For  a 
fixed  transmission  voltage  and  material  of 
conductors,  the  most  economic  size  of  conductor  will  be 
found  by  applying  Kelvin's  law  extended  to  include,  in 
addition  to  the  cost  of  conductors,  that  part  of  the  cost 
of  towers,  insulators,  line  construction,  phase  modi- 
fiers, etc.  which  increases  directly  with  the  cost  of  con- 
ductors.    Kelvin's  law  is  as  follows: — 

"The  most  economical  section  of  a  conductor  is 
that  which  makes  the  annual  cost  of  the  PR  losses  equal 
to  the  annual  interest  on  the  capital  cost  of  the  conduct- 
ing material  plus  the  necessary  annual  allowance  for 
depreciation".  Stated  another  way,  "The  annual  cost 
of  the  energy  wasted,  added  to  the  annual  allowance 
for  depreciation  and  interest  on  first  cost  shall  be  a  min- 
imum". 


*Tht  calculations  and  the  illustrations  in  this  article  were 
made  in  such  a  way  as  to  tx?  equally  suited  for  the  series  of 
articles  on  "Electrical  Characteristics  of  Transmission  Circuits" 
and  the  Superpower  Survey.  Figs.  69,  70,  72  and  75  and  Charts 
XXIII.  XXV  and  XXVII  appear  also  in  the  report  of  the  latter, 
which  is  printed  as  Pnifcsmmal  I'afcr  123  by  the  United  States 
Geological  Survey.  Similarly,  Charts  XXIV.  XXVI  and  XXVIII 
appear  in  the  Paper  bv  L.  E.  Imlav  in  the  Jnitrihil  of  the  A.  I. 
/■:.  /•;.  for  June.  1921.  (Ed.) 


FIG.    69 — THE    TR.ANSFORMEE    AND    CONDENSER     ARRANGEMENT     UPON     WHICH 
THE  CALCULATIONS   FOR  THE  220KV    PROBLEM    HAVE    BEEN    BASED. 

It  is  not  intended  that  this  arran(cment  would,  upon  a  complete  study 
of  the  problem,  be  found  to  be  the  most  desirable.  If  single-phase 
transformers  were  selected,  possibly  three  banks  for  each  double  circuit 
would  be  found  more  desirable  than  four  banks,  as  indicated  above. 


the  values  in  Table  Y  show  that  the  smallest  size  cable, 
605  000  circ.  mil.  will  be  the  cheapest  to  install.  At  0.3 
cents  per  kw-hr.  the  power  loss  for  this  cable,  capitalized 
at  15  percent,  represents  the  equivalent  of  an  investment 
of  $2  593  000  for  the  four  three-phase  circuits,  where- 
as the  cost  of  the  conductors  is  $3  224000.  If  the  cost 
of  power  loss  is  taken  as  0.4  cents  per  kw-hr.,  the  next 
larger  cable  will  be  the  most  economical  size  to  use, 
provided  that  there  is  no  increased  cost  of  towers,  in- 
sulators, etc.  If  the  losses  in  transformers  or  conden- 
sers vary  for  the  different  sizes  of  cables  compared 
such  losres  should  be  included  with  the  conductor  losses. 
There  is  always  a  question  as  to  what  price  should 
be  charged  in  Kelvin's  equation  in  estimating  the  cost 
of  power  loss.  If  all  power  saved  could  be  promptly 
sold,  the  cost  to  allow  might  be  considered  the  cost  at 
the  consumers  meter.  If,  on  the  contrary',  none  of  the 
power    saved  can    be   sold  under   any   circumstances, 


**An  interesting  graphic  presentation  of  Kelvin's  Law  is 
.civen  in  the  article  by  Mr.  L.  J.  Moore  in  the  Electrical  World 
for  Sept.  24,  1921,  p.  612. 


146 


A  TYPICAL  220  KV  PROBLEM 


then  the  cost  to  allow  is  the  cost  at  the  generating     (Charts  XVIII,  XIX,  XX  and  XXI).     When  using 
switchboard.     Intermediate  cases  may  occur.  charts  it  is  desirable  to  read  the  results  from  them,  at 

The  conductor  losses  of  Table  Y  were  taken  from     two  different  times  as  a  check  against  errors  in  reading, 


the  calculated  values  by  the  complete  method  A  listed 
in  Table  V*.     It  is  usually  sufficient  to  calculate  the 

TABLE   Y— APPLICATION  OF  KELVIN'S  LAW 


Conductors 
Circ.  Mill. 

Total  Loss 

in  12 

Conductors 

Kw 

Cost  of  Power  lost 
In  12  Conductors,  Capitalized  at  isJS 

Cost  of  12 
Conductors 

at  2QC 

per  Lb. 

At  0.3c 
per  Kw-hr. 

At  0.4c 
per  Kw-hr. 

At  0.5c 
per  Kw-hr. 

♦605  000 
715500 

18504 
15S40 

$2  593  coo 

$■2  220000 

|3  458  000 
$2  960000 

$\  322  000 
$3700000 

$3224000 
$3  837  000 

795000 
954000 

14304 
II  712 

|2  040000 
$1  641  COO 

$2  673  000 
$2  188  000 

$i  341  000 
$2  736  000 

$4  244000 
$50:1  000 

*Thi8  is  the  smallest  conductor  which  is,  in  this  case,  permissible 
on  account  of  corona  limitations.  These  tabulations  are  total  for  four 
three-phase  circuits.  It  will  usually  be  sufficiently  accurate  to  calcu- 
late the  conductor  I'^R  loss  for  one  size  of  conductor  and  assume 
that  the  loss  for  other  sizes  will  be  proportional  to  their  resistances. 
This  assumes  that  the  distribution  of  current  throughout  the  length 
of  circuit  will  be  approximately  the  same  for  the  different  sizes  of 
conductors  compared.  The  above  data  is  based  upon  75  000  kw  at 
85  percent  power-factor,  three-phase,  60  cycles,  delivered  over  each  of 
the  four  circuits  a  distance  of  225  miles  at  220  kv  with  a  50  000  kva 
condenser  in  parallel  with  the  load  on  each  of  the  four  circuits  and  an 
average  load  equivalent  to  80  per  cent  of  full  load.  It  should  be  noted 
that  the  third,  fourth  and  fifth  columns  do  not  give  the  actual  cost 
of  the  power  lost,  but  give  instead  the  values  at  which  these  losses 
are  capitalized. 

loss  in  the  conductors  for  one  size  of  cable  and  to  esti- 
mate it  for  other  sizes  of  cable,  assuming  that  this  loss 
varies  as  the  resistance  of  the  conductors,  that  is,  for 
a  given  line,  frequency,  load,  delivery  voltage  and  con- 
denser capacity  the  current  distribution  in  the  line  is 
approximately  the  same  for  various  sizes  of  conductors 
likely  to  be  considered.  Since  the  conductor  loss  varies 
as  the  square  of  the  current  and  directly  as 
the  resistance,  it  will  be  sufficient  to  estimate 


or  the  constants  may  be  read  from  both  the  Wilkinson 

and  Kennelly  charts  and  the  results  compared.     From 

Table  V  we  find  r  =  0.154  ohms,  so  that  R  =0.154  'X 

225  =  34.65  ohms  and  x  =  0.792  so  that  X  =  0.792  'X 

225  =  178.2  ohms.     From  Table  X  we  obtain  b  =  5.38 

X10-*  so  that  B  =:  5.38  X  225  X  io-^=  0.001211  mho. 

G  is  assumed  here  as  zero. 

From  Wilkinson  Charts  — 

fli  =  0.892 
and  since  rb  =  0.828 

32   =   0.020 

b\  =  32.2  ohms 
62  =  173-5  ohms 
and  since  r6^  =  4-457 

Ci  =  (too  small  to  read) 

C2    =   O.OOII75 

From  Dr.  Kennelly's  Charts — We  must  first  obtain 
the  hyperbolic  complex  angle  of  the  circuit  as  fol- 
lows : — 

Z=  34-65 +7178.1 
=  181-54    (78  59'46'' 

Y  =  o  +y  0.00121 1 
=  0.00121 1    190° 

ZY  =  0.21984  \i6£s9U§Z 

e  =  i/2r  =  0.4689  /84'29^53'^ 
Shift  e 

From  Chart  XIX,  — 7—  =  0.964  /o-4° 


From  Chart  XXI, 


0 

Tank  6 
6 


0.964  /d'lAr'oa'' 


=  1.0785  \o.88" 
=  1.0785  \o"52'48" 


TABLE    Z— CABLE    AND   CIRCUIT  CONSTANTS    CORRESPONDING 
TO  A  THREE-PHASE,  60  CYCLE    CIRCUIT,    225    MILES    LONG 
the   loss   for   other   conductors   as   being   in-  CONSISTING  OF  FOUR  SIZES  OF  ALUMINUM   CABLES  OF  AN 

versely  proportional  to  their  resistance.  ARRANGEMENT  EQUIVALENT  TO  21  FEET  DELTA 

The  various  constants  corresponding  to 
the  four  sizes  of  conductors  considered  are 
listed  in  Table  Z.  It  may  be  interesting  to 
note  the  variation  in  these  constants  corre- 
sponding to  the  different  sizes  of  cable  for 
the  high-tension  line  alone,  and  also  when  the 
transformer  impedances  are  included  with  the 
line  impedance. 


SOLUTION  OF  THE  220  KV  PROBLEM 

Assuming  that  605  000  circ.  mil.  alumi- 
num-steel cables  work  out  as  the  most  econo- 
mical size,  the  next  step  is  the  determination 
of  the  auxiliary  constants  A,  B,  and  C  for 
this  size  of  conductor,  spacing  and  60  cycles. 
(These  constants  would  have  previously  been 
determined  when  determining  the  most  eco- 
nomical size).  Mathematically  these  constants 
may  be  calculated  by  real  hyperbolic  func- 
tions (Chart  XVI)  or  by  convergent 
series  (Chart  XI).  Graphically,  they  may  be  ob- 
tained from  Wilkinson's  charts  (Charts  V,  VI  and 
VII)  or  through  the  medium  of  Dr.  Kennelly's  charts 


AREA  Of  CONDUCTORS  (C  M  p 

DIA 
OF 

ALUM 
COND. 

STRANDS 

LINEAR  CONSTANTS  OF  LINE 
TO  NEUTRAL 

IMPEDANCE  TO  NEUTRAL        1 
OF  SO  000  KVA  BANK  OF        | 

ALUM 

BTttL 

TOTAL 

COPPER 

EQUIV 

AL 

ST 

r 

X 

g 

.\. 

R 

X 

G 

B„ 

■  10-8 

R,» 

Xr. 

Gtn 

^i;-. 

(toS.coc 

7B.00O 

68  3,iC0 

i-fj^j,^ 

0.9  SI 

S* 

' 

0./34 

0.7  •}! 

0 

6.38 

J-S.*! 

/  '*..- 

0 

IZ<' 

4.37 

'f.it 

o 

0 

y'i.sQo 

<ii.foo 

eoa.'9oo 

AiO.OOO 

,..,. 

S-* 

7 

0./3' 

0.792 

0 

S.fS 

:  9.-9  8 

I7S.9 

0 

'234, 

«.3' 

7  9.t4 

0 

o 

793,000 

103. lOQ 

BtS.ioo 

eg^.^ 

I.09Z 

J-* 

7 

o.'<; 

0,TS 

0 

S.-9  9 

ii,33 

'7t.9- 

0 

/93S 

i.-i'' 

>9.6  9 

0 

o 

96-^.000 

'i3.7O0 

',07  7.  700 

ioo.ooo 

i.i9(> 

S4 

7 

0.097g 

O.7.. 

0 

S.58- 

2  3.00 

'7 '.9 

0 

/2S6 

i.3r 

.■«.*■# 

o 

c 

UNEAH  CONST AhfTS 

HYPERBOLIC  QUANTITIES 

AUXILIARY  CONSTANTS  OF  CiRCUiT 

R  1  X  [GLS. 

e-Yz^           1          zo-yi 

(A)              1               (B)             1              (C) 

♦                                                                                               MI0HTlM(OI(l.lN[,tOHCUT«*ii                                                                                                                                                                                                          | 

i^QS.OCQ 

^■*^<, 

'7g3 

" 

>3,> 

J8 

.^93->SS*3.C3o:>3', 

.•-:;;j,-:^' 

~-'°cV.!'^°*\9o-T*''2*-f' 

7.2\5'J0'C7- 

7/ 5.  SCO 

if-iS 

ItSt 

0 

IZ34, 

.02g,9^J.-,t.99 
=  .th7b/9S''->    =11 

3a 

=  in.oilBa'so'"" 

-.cao  DflJ^j.OO-.Si 

.'t  W*-)5'i5' 

79S.OOC 

Xi33 

17^4 

c 

>Z:>S 

-.^it7/S^^93-Zf 

37 

=  ,ffS,4//-0'/t- 

.',ycV.  I'sV^v^y 

::T.-VA9i=f8-s:' 

9  6 -f.  000 

Z300 

mf 

" 

/SS6 

■  039£»-f*j-'*tS'6 
^.Aht.)  isJ  3r,z' 

J7 

■  f939'*t  1  j.O/ai3« 

.'°oo.°°!'(2^^' 

(.OS.OOO 

J?83 

3itia 

0 

IZ:l 

.0444. 7*3.5. S?f? 

■♦3 

3*.Sii3^jJ09..J 
^3,'.(.7jBC-3i-0i- 

=  .O0.i5«V.'0-Jiil- 

7.-t7\-)'53'y«* 

TS.SOO 

3S  6i 

3,S?3 

0 

ilti 

=..87079  II'  'i"3f,' 

31.837  fjZOt.SfS 

=  ZOI.>.9/i''9t.--1- 

(f/*" 

-.^^..,X\9o-iVi*' 

79S.COO 

i9si 

3M31 

0 

laiS 

.03S-i*7rj.S'SSf 

4'S.-»-#\.}*JiVJ" 

Ji..*j-fi-»j_jo;.ojfl* 

-,.e>o,.ai\9o-3-'IS' 

9S^.0C0 

3S  19 

3..U 

° 

l3Si 

4/J 

S3:.1fl->-j3oa.SBB 

-ao-i.99lsV3rso' 

-  ,oo  "01  \90'mi* 

o/\j-a3-?s' 

bOS.OOC 

',102 

IS7  » 

0 

,3n 

■*«. 

.f^''f=*J.033Sit- 
:.89S'  fii    31- 

3i.»l-*U-'f.93^ 
-.Z47.l.7/8rZi-3r 

-.noii99\90-2ro3' 

.3W3.-..' 

7'S.SOO 

3S'9S 

tsss 

0 

/13(, 

.O39'Jt3.St/03 
s:..S42-f  /8fOO^ 

■*s 

.8f  73i-*-j.oaoff-»4 

^.a-Ji  /''39'3-il 

=  3^4.8  ihu£i.t' 

~J.OO"l.-i\9'o''s'i-'*3- 

79S.0O0 

32  7C 

3SA 

0 

1X3^ 

=  .St3t/g*-'9-SS' 

^i5.1^3--fO'<J-*' 

,  241.9  jsroyzs' 

-.oo,.7/l^o-?J-i»" 

9S^.0DC 

...T 

ZSI  s 

0 

""■ 

-:°l'ifVi*'-vV'iJ'' 

-f-*8.9Fl3*/3'oa 

^.8tfi/rc8-3r 

^^.,0/S3'J-p4  3' 

~.000  C07t  ].00'l9/ 

..oo"f'\:io:3c:53i 

♦Since  two  50000  kv-a  banks  of  transformers  will  be  required  at 
each  end  the  corresponding  values  tor  impedance  will  be  half  these 
amounts. 

Sinhdie      0.964      h^M'^o" 


♦In  the  Journal  for  Dec.  1921,  p.  544. 


Tankdie      1.0785    \o''52'48" 
=  ^0.8939  A°i6^48^" 
a,  =  0.8937 

fl2  ■=    0.01096 


A  TVPICAl.  220  KV  PROBLEM 


147 


B  -  Z^^^  -  181.54  /78°59'46"  X  0.964  /d'u'oo" 
0 

—  175.0  l-jifi3'^6"  ohms 

b\  —  31.2  ohms 
dj  »  172  ohms 

C-  K =  0.001211    [go^  X  0.964  ^.^i^/ca" 


=  0.001167  \  90°24''oo^^  mho 
O  =  —  0.000008  mho 
f2  ■■  o. 001 167  mho 

The  auxiliary  constants  as  obtained  graphically 
and  by  exact  mathematical  solution,  are  given  in  Table 
ZZ.  It  is  thus  seen  that  the  Kennelly  charts,  although 
primarily  intended  for  correcting  the  linear  impedance 
and  the  linear  admittance  of  circuits  for  the  equivalent 
IT  solution,  are  highly  adaptable  to  determining  the 
values  of  the  auxiliary  constants  to  a  very  close  degree 
of  accuracy.  The  use  of  these  charts  for  obtaining 
auxiliary  constants  requires  more  arithmetical  work 
than  the  use  of  the  Wilkinson  charts.  For  instance 
the  hybolic  angle,  Q  =  yzT  of  the  circuit  must  first 
be  calculated  before  the  charts  can  be  employed.  The  re- 
sults, read  from  charts,  must  then  be  multiplied  by  the 
impedance  and  the  admittance  of  the  circuit  for  obtain- 
ing auxiliary  constants  B  and  C.  Auxiliary  constant 
A  cannot  be  taken  directly  from  a  single  Kennelly  chart. 
To  obtain  this  auxiliary  constant  from  these  charts  it 
is  necessary  to  divide  the  values  read  from  two  of  these 

sinh  6/e 

charts  since  A=   - — .     „  , .    .     Chart  tanh  6/6  is  con- 
tanh  6/6 

structed  for  angles  up  to  and  including  0.50  polar 
values.  This  makes  it  adapted  to  angles  up  to  i.O  polar 
value  when  used  for  determining  correcting  factors  for 
the  equivalent  tt  solution.  This  is  for  the  reason  that 
for  obtaining  such  correcting  factors  we  enter  this  chart 
with  6/^.  However  for  obtaining  auxiliary  constant 
A  by  means  of  values  read  from  these  charts  we  must 
enter  this  chart  with  6  in  place  of  6/2.  This  limits  the 
use  of  the  Kennelly  charts  for  obtaining  auxiliary  con- 
stant A  to  circuit  angles  not  exceeding  0.5  polar  values. 
In  case  the  circuit  angle  has  a  polar  value  greater  than 
0.5,  Wilkinson  chart  A  may  be  used  provided  the  line 
is  not  over  300  miles  long.  If  the  circuit  is  over  300 
miles  long  the  auxiliary  constants  should  be  determined 
by  mathematical  calculation. 

In  the  following  discussion  the  calculated  values 
for  the  auxiliary  constants  will  be  used,  since  exact  re- 
sults are  required  for  the  purpose  of  comparing  the  re- 
sults with  those  obtained  by  the  approximate  method, 
a  description  of  which  follows  the  complete  solution. 

NORMAL  LOAD — COMPLETE   SOLUTION 

The  complete  solution  for  normal  load  is  given  by 
Chart  XXIII.  At  the  top  is  illustrated  the  circuit  dia- 
gramatically.  Underneath  this  is  stated  the  load  con- 
ditions, linear  and  the  auxiliary  constants  for  this  cir- 
cuit. The  transformer  data  and  method  of  determining 
the  amperes  iron  loss,  magnetizing  current  and  im- 
pedance to  the  neutral  of  the  lowering  transformer  is 


also  shown.  Actually  the  impedance  of  raising  and 
lowering  transformers,  even  when  duplicates,  is  slight- 
ly different  when  the  connections  are  not  made  to  simi- 
lar taps.  This  difference  is  so  slight  (and  so  far  as  the 
raising  transformer  is  concerned  so  unimportant)  that 
for  simplicity,  we  are  assuming  that  both  raising  and 
lowering  transformers  have  the  same  impedance.  This 
comprises  all  the  data  required  for  a  complete  mathe- 
matical or  graphical  solution  of  this  circuit. 

Following  the  data  is  a  complete  graphical  vector 
solution  of  this  circuit  with  symbols  placed  on  all  vec- 
tors indicating  the  manner  of  obtaining  their  values. 
At  the  lower  left  hand  corner  is  placed  a  complete 
mathematical  solution  of  the  problem,  which  parallels 
the  graphical  solution  (one  method  of  solution  check- 
ing the  other).  In  the  calculations  of  the  high-voltage 
circuit  the  curretjt,  in  order  to  include  the  power-fac- 
tor, must  always  be  expressed  in  complex  form  referred 
to  the  vector  of  reference,  as  indicated  by  a  dot  under 
the  symbol  I. 

At  the  lower  right  hand  comer  a  method  is  indi- 
cated of  determining  the  transmission  loss  from  the  cal- 
culated  quantities.     The  loss   in   the  high-tension   line 

TABLE  ZZ— AUXILIARY  CONSTANTS  FOR  220  KV 
PROBLEM  APPROXIMATE  SOLUTION 


Calculated 

From 
Wilkinson  Chart 

I-"rom 
Kennelly  Chart 

ai 

Ci 

c= 

0.893955  =  100% 
0.020234  —  100% 
32.198=100% 
172.094^  100% 
-0.000008=100% 
0.001168=100% 

0.892  =  9978% 
0.020  =  98.85  % 
32.2=  100% 
1735=  100.82% 

can't  read 
0.001175  =  100.6% 

0.8937  =  9997% 
0.01996  =  98.65  % 
32.2=100% 
172  =  99.95% 
-0.000008=100% 

o.ooi  167  =  99.91  % 

can  be  determined  graphically  by  scaling  off  the  voltage 
and  the  current  at  each  end  of  the  high-tension  line  and 
measuring  the  angle  between  the  vectors  representing 
the  current  and  the  voltage.  The  current  times  the 
voltage  times  the  cosine  of  this  angle  will  give  the  power 
at  the  point  considered  and  the  difference  between  the 
power  as  so  determined  at  the  two  ends  of  the  high- 
tension  line  is  the  line  loss.  The  losses  in  transformers 
and  condensers  are  known  and  stated  at  the  top  of  the 
chart. 

The  complete  vector  diagram  is  constructed  as  fol- 
lows :  First  draw  the  horizontal  line  representing  Ei.s, 
the  voltage  at  the  load  to  neutral.  This  should  be 
drawn  to  as  large  a  scale  as  possible.  All  other  voltage 
vectors  will  of  course  be  drawn  to  the  same  scale.  The 
vector  /l  representing  the  load  current  is  now  drawn 
to  as  large  a  scale  as  can  be  used  without  mixing  the 
current  vectors  with  the  voltage  vectors.  This  is  drawn 
at  an  angle  of  31°  47'  from  £ln  in  the  lagging  direction, 
corresponding  to  a  lagging  load  of  85  percent  power- 
factor.  It  usually  works  out  that  for  normal  load  the 
power-factor  at  the  receiving  end  should  be  slightly 
lagging  and  at  the  sending  end  slightly  leading  so  that 
the  average  power-factor  of  the  line  will  be  close  to 
unity.  This  will  necessitate  a  phase  modifier  in  paral- 
lel with  the  load,  having  approximately  the  capacity  of 
the  lagging  kv-a  in  the  load. 


148 


A  TYPICAL  220  KV  PROBLEM 


The  lagging  kv-a  in  the  load  is  equal  to  the  kv-a  of 
the  load  times  the  sine  of  the  angle  of  the  load.  In  this 
case  it  is  88235  X  sin  31°  47'  =  46500  kv-a.  The 
vector  diagram  is  constructed  on  the  basis  of  a  45  000 
kv-a  condenser  in  parallel  with  the  load.  This  con- 
denser has  a  power  loss  of  4.72  amperes  to  neutral  and 
since  this  is  in  phase  with  the  load  voltage,  we  trace  from 
the  end  of  the  load  current  vector  horizontally  to  the 
right  a  distance  representing  4.72  amperes  by  the  cur- 
rent scale.  The  current  per  terminal  for  the  condenser  is 
118.09  amperes  so  that  the  leading  component  of  the  cur- 
rent input  of  the  condenser  is  118.00  amperes.  Since  this 
is  leading  it  is  drawn  vertically  upward  from  the  last 
point  determined.  Actually  we  will  not  need  to  deter- 
mine the  1 18  amperes  leading  component,  but  will  com- 
plete the  solid  black  condenser  triangle,  since  the  length 
of  the  input  line  is  1 18.09  amperes.  To  the  vector  sum 
of  load  and  condenser  currents  thus  determined  we  now 
add  the  leakage  current  of  the  lowering  transformers, 
the  lagging  component  of  which  materially  effects  the 
capacity  of  the  phase  modifiers  required  because  of 
its  nearly  direct  opposition  to  it  under  load.  We  have 
assumed  that  the  leakage  current  required  by  the  low- 
ering transformers  will  be  supplied  by  the  phase  modi- 
fier on  account  of  its  close  electrical  proximity  to  the 
lowering  transformers.  On  this  assumption  the  trian- 
gle representing  this  transformer  leakage  will  be  located 
as  indicated.  There  is  a  loss  current  of  1.85  amperes  in 
phase  with  the  load  voltage  and  a  magnetizing  current 
of  13.9  amperes  in  lagging  quadrature  with  the  load 
voltage.  We  thus  find  that  the  current  /r  at  the  receiv- 
ing end  of  the  line  is  204.17  amperes,  lagging  5°  i'  16" 
behind  the  load  voltage.  In  this  case  the  magnetizing 
current  of  the  lowering  transformer  reduces  the  effec- 
tive capacity  of  the  phase  modifier  by  an  amount  of 
13.9  amperes ;  that  is  by  5.3  percent  of  the  total  capa- 
city of  the  lowering  transformers. 

We  next  determine  the  voltage  at  the  high-voltage 
side  of  the  lowering  transformers;  that  is  the  voltage 
jETrn  at  the  receiving  end  of  the  transmission  line. 
Knowing  the  resistance  and  reactance  of  the  lowering 
transformer  banks  to  neutral  and  the  curent  /r,  the 
transformer  resistance  voltage  drop  is  plotted  in  phase 
with  the  current  /r  and  the  reactance  voltage  drop  in 
quadrature  with  the  resistance  drop  as  indicated.  The 
voltage  at  the  sending  end  £sn  of  the  transmission  line 
is  next  determined  by  applying  auxiliary  constants  A 
and  B  to  the  voltage  and  current  respectively  of  the  re- 
ceiving end. 

The  base  of  the  impedance  triangle  for  the  high- 
tension  line  In  X  ^1  represents  the  resistance  drop  of 
the  high-tension  line  in  phase  with  the  receiving  end 
current.  In  quadrature  to  this  is  the  reactance  volts 
drop  of  the  line  /r  X  ^2-  The  voltage  at  the  sending 
end  is  thus  determined  to  be  131  858  volts  which  cor- 
responds to  slightly  less  than  230  000  volts  between  con- 
ductors. An  arc  of  a  circle  corresponding  to  the  volt- 
age to  be  maintained  at  the  sending  end  will  serve  as 


a  guide  in  determining  the  proper  capacity  condenser 
necessary  to  maintain  this  sending  end  voltage.  An  in- 
crease in  condenser  capacity  rotates  the  vector  /r  in 
a  counter-clockwise  direction,  swinging  the  line  im- 
pedance triangle  also  in  a  counter-clockwise  direction 
thus  decreasing  the  voltage  £sn  and  reducing  the  line 
drop.  A  decrease  in  condenser  capacity  rotates  the 
vector  /r  in  a  clockwise  direction,  swinging  the  line 
impedance  triangle  also  in  a  clockwise  direction,  thus 
increasing  the  voltage  £sn  and  increasing  the  line  drop. 
Thus  the  effect  upon  line  voltage  drop  may  be  readily 
determined  for  condensers  of  various  capacities. 

The  next  step  is  to  determine  the  current  at  the 
sending  end.  This  is  done  by  applying  auxiliary  con- 
tants  A  and  C  to  the  current  and  voltage  respectively 
of  the  receiving  end.  It  will  be  noted  that  the  charg- 
ing current  is  drawn  as  leading  by  90  degrees  the  high- 
tension  voltage  at  the  receiving  end,  which  voltage  is 
taken  as  the  vector  of  reference  as  in  previous  discus- 
sions. The  current  at  the  sending  end  is  thus  deter- 
mined to  be  220.34  amperes  leading  the  vector  of  ref- 
erence by  35°  12'.  The  impedance  triangle  for  the  rais- 
ing transformers  may  now  be  drawn  in,  the  resistance 
drop  of  same  being  drawn  parallel  with  /g.  This  then 
gives  the  voltage  at  the  generators.  The  current  at  the 
generators  is  determined  by  adding  vectorally  to  /g  the 
leakage  of  the  raising  transformers.  It  is  assumed  that 
the  raising  transformers  will  receive  their  excitation 
from  the  generators,  in  which  case  the  leakage  trian- 
gle will  occupy  the  position  shown,  resulting  in  a  cur- 
rent at  the  generators  of  218.88  amperes. 

NORMAL   LOAD — APPROXIMATE    SOLUTION 

The  approximate  solution  for  normal  load  is  given 
in  Chart  XXIV.  It  differs  from  the  complete  solu- 
tion in  that  the  impedance  of  the  lowering  transform- 
ers is  added  to  and  considered  as  a  part  of  the  line  im- 
pedance so  that  there  are  no  transformer  impedance 
triangles  to  construct.  It  differs  also  in  that,  in  the 
case  illustrated,  the  conditions  at  the  sending  end  only 
are  obtained,  whereas  in  the  complete  diagram  the  con- 
ditions at  both  sending  end  and  generators  were  deter- 
mined. If  the  condition  at  the  generators  in  place  of 
at  the  sending  end  is  required,  the  impedance  of  the 
raising  transformers  would  also  be  added  to  that  of 
the  line,  the  general  construction  of  the  diagram  remain- 
ing the  same  as  for  the  complete  solution. 

If  it  is  not  necessary  to  know  conditions  at  both 
sides  of  the  raising  and  lowering  transformer  banks, 
then  it  will  be  seen  from  a  comparison  of  the  two  dia- 
grams that  the  approximate  solution  will  be  simpler,  al- 
though the  results  will  be  somewhat  incorrect.  For  in- 
stance, for  the  220  kv  problem  illustrated,  the  errors 
in  the  results  will,  according  to  tabulations  in  the  lower 
right  hand  corner,  vary  from  0.88  to  2.38  percent.  If 
the  losses  in  condensers  and  transformers  were  not 
added  to  the  load  (as  they  are  in  both  these  complete 
and  approximate  methods)   and  the  transformer  mag- 


A  TYPICAL  220  KV  PROBLEM 


i» 


CHART   XXIII— 220   KV   PROBLEM— NORMAL    LOAD 


(COMPLETE  SOLUTION) 

■<LOW  TENSION  VALUES  REFERRED  TO  THE  HIQM  TENSION  OlROUITl 


RAISING  TRANSFORMERS 
Ztn~ 3' 186^3  39. 62  OHMS 


HIGH  TENSION  UNE 


"Z  ■  34.66  ♦)  I  78.2  OHMS 
Y-     0        »i  .001211    MHO 


LOWERINO  TRANSFORMERS 
."Zjft  -  3  1 86  *!  39.82  OHMS 

/<L*  '0_ 


rr^f-]  [  T  T  T  I  1  I  1  1  I  T  T  y  T  T  iT-"^1 


SENDING  END 

NORMAL  LOAD 


PER  3  PHASE 

PER  PHASE 

ciRbuiT 

TO  NEUTRAL 

KV-AL-e6  238 

KV-Aln-"^i» 

KW|  •  76.000 

Kvy/LN-"""" 

PF,_-  86%  LAG. 

PFln-'""^°- 

E|_"  220.000 

Eln-i""" 

1|_-23l  66 

IlN-231  66 

eo  OVOLES 

CONDENSER 

(ONE  REQUIRED) 

3  PHASE 

TO  NEUTRAL 

NEUTRAL 

LINEAR  CONSTANTS 

Z  -  34.86  ^j  I  78.2  OHMS 
Y  "    0       +j  .001211    MHO 

AUXILIARY  CONSTANTS 

(a)  ■  (3| » j  82)  -  COSH  e  -  893.965  *j  020.234  -  e94io/rir4r 

(B).(b|«,Bj).zS!!|l«.  fIsiNHe.32  198  4^172.064  OHMS 
(C)-(O|<-»02)-lfSi^-=l=8INH8--.000.00»»j.00l.ie8MHO 
WHERE  e-V^  I* 


RCOEnmacNO 


TRANSFORMERS 


rrVW)  BANKS  IN  PARALLEL  AT  EACH  END  OF  THE  LINE) 

I  RESISTANCE  VOLTS  -  0  668  * 

'  REACTANCE  VOLTS  -  8  226  » 

1  MAGNETIZING  CURRENT     -  6.3O0  % 
[iron  LOSS -0.706% 

VALUES  TO  NELITRAL. 
KV-Ato"33.333.  Etn"'"""  lT„-SeS.4 


ON  BASIS  OF 

100.000  KV-A  RATING 

FOR  TWO  BANKS 


-TN 


'TN- 


Xtn' 


.00868  » 127.020 

282.4 
.08226  it  1 27.020 

282.4 


-  3. 1 86  OHMS  RESISTANCE 

-  39.82  OHMS  REACTANCE 


MAGNETIZING  CURRENT  -  ""lay";"^'"  -  1 3.9  AMPS  AT  1 27.020  VOLTS 


KV-Ag-  46.000        K  V-Agu  =  1 6.000 

Eq- 220.000  Eon"  1 27.020 

lQ-na.09  Iqn-iis.os 

KWc-.n«-l600  KWo-Loaa-N-600 
In 


IRON  LOSS 
IRON  LOSS 


.,-4.72 


•O-CO88-N 
■C-LOSS"*"  'o-L03a-N"<" 

NOTE  -  THE  CONDENSER  INDICATED  BY  BROKEN 
UNE  CIRCLE  SERVES  AS  A  SPARE  DURING  NORMAL 
OPERATION  BUT  IS  REQUIRED  FOR  THE  EMERGENCY 
CONDITION. 


LINE    CHARACTERISTICS 

LENGTH  OF  TRANSMISSION  226  MILES 

CONDUCTORS-  3-606.000  CM  ST.  REINFORCED  ALUMINUM 

DIAMETER  OF  CONDUCTORS  963- 

MAXIMUM  ELEVATION  600  FEET 

SPACING-  EQUIVALENT  10  21'  DELTA  SPACING' 


IMPCOANCC  Of 

RAISING 
riUNSFOAMCMa 


-m  1.B6  AMPS  AT  1 27.020  VOLTS 


,.00706  "33. 333.333 
127.020 
.         -  236  KW  TO  NEUTRAL 
NOTE-FROM  THE  VECTOR  DIAGRAM  IT  MAY  BE  SEEN  THAT  THE  RAISINO 
TRANSFORMERS  WILL  BE  EXCITED  BY  A  VOLTAGE  EQUIVALENT  TO  129.918- 
THE  LEAKAGE  CURRENT  REQUIRED  TO  EXCITE  THE  RAISING  TRANSF0RMCR8 
VfILL  THEREFORE  BE 
\  MAGNETIZING  CURRENT  -13.8  AMPS.  AT  1 28.8 1 6  VOLTS 

IRON  LOSS  -  1 .6 1  AMPS.  AT  1 29.9 1 8  VOLTS 


CALCULATION 
FOR   RECEIVING-END 
CURRENT  AND  VOLTAGE 

I|_-l96.82-j  121.97  AMPS.  TO  VECTOR  £, 

•o' 

Ij-       1.88 
lo- 203.39 


4.72«jllS.OO 
1.88-1    13.90 


COS  6*01    I6'«.996I8 

SIN  6*  or  18' -.087523 

COS  8'3e' l9'-.98e74 

SIN  8*38'  19*-   14983 

-204.17     \6'  0 1  ■  1 8*  AMPS.  TO  VECTOR    E.^^, 

-  204.1  7  \8*  36  1 8'  AMPS.  TO  VECTOR  OF  REFERENCE 

"i 


•201.87 -i    30.66 


E(,N  "  Yd  27.020  «  998I8*204.I7»3.I86)'*<I27.020«  087623*204.17  «  39.82)* 
-       128.630  /8'38'  1 9*  VOLTS  TO  VECTOR   If, 

CALCULATION    FOR    HIGH   TENSION    CIRCUIT 


■  Il4.999»j    2.803 
I.767»j33.767 


!r(B)- 


EsN*'3S'7*>*i38.38ff 

-  1 3 1.86S  /|8*00'24*  VOLTS 


Jr(A)-I8I.08-j    23.23 

Epf|(C)-  -l.03*i  160.24 

l3-IB0.06*jl27.0l 

-  220.34  /36*I2  00*  AMPft 


DETERMINATION   OF  LOSSES 

KWln  -    26.000 

I^WrN  -  1 28.830  «  304. 1 7  1003    8' 98' I  r)  -  26.988 

KWg^  -  13l.86a«220.34(OO319*  1 1' 38-)  -  27.439 

KW(jEN-N"  I29.9I6«2I8.88  (00811*62' 28*)  -    27.827 

LOSSES  TO  NEUTRAL 

LOWERING  TRANSFORMERS  AND  CONDENSER 

26.988-28.000  -      888 

HIGH  TENSION  LINE  27.430 -26.088  -     1471 

RAISINO  TRANSFORMERS  27.827  -  27.439  .       888 

TOTAL  LOSS  TO  NEUTRAL  -    2827 

AS  A  PARTIAL  CHECK 

LOWERING  TRANSFORMERS 

(IRON  LOSS  1. 86 XI 27.02    ^-       9SB 

(COPPER  LOSS  204.1 7'  x  3. 1 e«)  -        132 

SYNCHRONOUS  CONDENSER      ■       SOO 

RAISINO  TRANSFORMERS 

(IRON  LOSS  1.8  I  «  129.918)' ■        19S 

(COPPER  LOSS  220.342x3  186)  -         184 

HIGH  TENSION  LINE 

(VALUE  ABOVE  ASSUMED  AS  CORRECT  -      1471 
2827 

1 8.00  AMPS 

EFFICIENCY 


CALCULATION  FOR  GENERATOR  VOLTAGE  AND  CURRENT 


EFFICIENCY.  (HIGH  TENSION  UNE) 
EFFICIENCY.  (GENERATORS  TO  LOAD) ' 


g»  966  ^  ».,.- 
27.439  •*"» 
26.000 
27.827 


'  89  84* 


Eqen^-V  (I3I.868X  94442  +  220.34x3. I86)2*(I3I  868  x  32878-220.34x39  82)2 
-  1 29.0 1 6  /l  6*  26  03*  VOLTS  TO  VECTOR  Ig 
•  128.918  /l9*466r  VOLTS 

K V-Ag  "  1 3 1 .  868  X  220  34  X  3  -  87. 1 83  PER  3  PHASE  CIRCUIT 

I^V-Agj^-  I  29.9 18x21  8  88  x  3  -  86.308  PER  3  PHASE  CIROOIT 

l^y  ./\  QQ-I3I868X|60.24«3-  69.43 1  PER  3  PHASE  CIRCUIT 


220.34  (.98394  *  j  .288 1 3) 
-21 2.39  ♦i  58.84 
.       1.81 -j  13  80  (LIAIOUie  OF  RAISINO  TRANSFORMERS) 

'0EN-N-"*"*i"°* 

-  2 1 6.88  /ir62'  28*  AMPERES  TO  VECTOR  Eqe^ 
•  318.88  /3r  38' 26*  AMPERES 


150 


A   TYPICAL  220  KV  PROBLEM 


CHART   XXIV— 220    KV   PROBLEM— NORMAL   LOAD 

(APPROXIMATE  SOLUTION) 

THIS  APPnOXIMATC  SOLUTION  ASSUMES  THAT  THE  IMPEDANCE  OF  THE  LOWERING  TRANsroflMERS  MAY  BC  ADDED  TO  THE  LINE  IMPEDANCE  AND  TREATED  AS  THOUGH  IT  WERE  DtSTRIBUTCO  LINE  IM< 
P£DANCE--THIS  ASSUMPTION  SIMPLIFIES  THE  SOLUTION  AT  THE  EXPENSE  OF  ACCURACY  (SEE  LOWER  RIGHT  HAND  CORNER  OF  PAGE;  ALSO  TEXTl.-THE  SOLUTION  BELOW  IS  BASED  UPON  THE  VOLTAGE  BEING 
HELD  CONSTANT  AT  THE  LOAD  SIDE  OF  THE  LOWERING  TRANSFORMERS  AND  AT  THE  MICH  TENSION  SIDE  OF  THE  RAISING  TRANSFORMCRS.-'IF  THE  VOLTAGE  IS  TO  BE  HELD  CONSTANT  AT  THE  GENERATOR  BUS, 
THE  IMPEDANCE  OF  THE  RAISING  TRANSFORMERS  MUST  ALSO  BE  ADDED  TO  THAT  OF  THE  UNE-ALL  LOW  TENSION  VALUES  ARE  REFERRED  TO  THE  HIGH  TENSION  CIRCUIT, 


RAISING  TRANSFORMERS 
Z-,.  =  3, 1  SB +  i  39, 82  OHMS 


HIGH  TENSION  LINE 


Z  = 


34,B6«i  I  78.2  OHMS 
0       -fj  .001211    MHO 


rr^^^^^ I »»iii)M I '^^^''^TTraiiCOTp^'^^ i  »»»M»p^^^ I »»i)Mi)|''^^^iiWMiiT^^'^^  I CMMa I  '■"■'VApnfjim^w^ 


'T    T    T 


U. 


LOWERING  TRANSFORMERS 
Ztn'^'SS^J'^  820HMS 

zikjo 


I    I    I    T    I    I    I 


SENDING  END 

NORMAL   LOAD 

PER  3  PHASE  PER  PHA 


RECEIVING  END 


CIRCUIT 


TO  NEUTRAL 


KV-Al-  88.236        K V-Aln  =  29.4 1  2 

K  Wl=  76.000  K  Wln  ■  26.000 

PF|_=86%LAQ.         PFln=B6%LAG. 

El=  220.000  Eln  =  1 27.020 

1^=231.66  Iln=23I.66 

60  CYCLES 

CONDENSER 

(ONE  REQUIRED! 
3  PHASE  TO  NEUTRAL 

KV-A(3N=I6.000 
~  127.020 


LINEAR  CONSTANTS 

Z  "  37.836  +  j  2 1  8.02  OHMS  * 
Y  =       0      +j  .001  21 1  MHO 
•k  THIS  INCLUDES  IMPEDANCE  OF  LOWERING  TRANSFORMERS 

AUXILIARY  CONSTANTS 

(a)  "  (3|  +  j  82)  "  COSH  a  =  .870783  -»  j  ,02 1 9 1  I 

(B)  =  (b|  t  j  bj)  .  z  ^"^"°  =  ^ 8INH  0  -  34,6663  ♦  j  208.83  OHMS 

(C)  -  (C|  +  j  C2)  =  Y  5!^  -  ^  SINH  9  —  .000.009  tj  .00 1 1 68  MHO 
WHERE  e-  yZY  l? 


TRANSFORMERS 

(TWO  BANKS  IN  PARALLEL  AT  EACH  END  OF  THE  LINE) 


ON  BASIS  OF 

1 00.000  KVA  RATING 

FOR  TWO  BANKS 


RESISTANCE  VOLTS        .  -0.668% 

REACTANCE  VOLTS  =8,225% 

MAGNETIZING  CURRENT  =6,300% 

IRONLOSS   =0,706% 


KV-A 
R 
X 


VALUES  TO  NELITRAL 
,^^-33.333.  E.^jj=  127.020  Itn=262.4 

00668X127,020 


KV-Ao'^ooo 
Eq=  220.000 


'TN  262.4 

.08226  X  I  27.020 
'TN  262,4 

,0630X33,333,333 
27,020 

IRONLOSS .00706X33,333,333, 

IRON  LOSS •  836  KW  TO  NEUTRAL 


MAGNETIZING  CURRENT  =  - 


3,186  OHMS  RESISTANCE 

39,82  OHMS  REACTANCE 

13.9  AMPS  AT  1 27,020  VOLTS 
:  1 ,86  AMPS  AT  1 27.020  VOLTS 


-ON 


KWo-LO38-l800   KWo_LOSS-N-«<'0 
ln-ins.<!='>-72  lr'_ins.o-N"*-72 


NOTE  -  THE  CONDENSER  INDICATED  BY  BROKEN 
LINE  CIRCLE  SERVES  AS  A  SPARE  DURING  NORMAL 
OPERATION  BUT  IS  REQUIRED  FOR  THE  EMERGENCY 
CONDITION. 


LINE   CHARACTERISTICS 

LENGTH  OF  TRANSMISSION  226  MILES 

CONDUCTORS-  3-606,000  CM  ST,  REINFORCED  ALUMINUM 
DIAMETER  OF  (XINDUCTORS  963-  rfS- 

MAXIMUM  ELEVATION  600  FEET  ,  !#    " 

SPACING- EQUIVALENT  TO  2  r  DELTA  SPADING 


CALCULATION   FOR 
RECEIVING-END  CURRENT 

,    -■l,9e.82-j  1 2 1 .97  AMPS.  TO  VECTOR  Ern 
|(j-      4.72  tj  1 18.00 

•t- 

lp-203.39-j    17.87 

=  204.17    \6'or  16'  AMPERES 

CALCULATION    FOR   HIGH  TENSION   CIRCUIT 

E      (A)-ll0.607-Fj     2.783  Id(A)  =  I  77,60 -(     11,11 


Id(B)=     I  0.762 -f  141,666 
•  " 

Es((=l2l,3e9+j44,e39 

-129.318  /20'ir36'  VOLTS 


„(C)°    -ll4Ti  147,09 
=  176,36*1  136,98 


=  222.69  /3r  Se'Of  AMPERES 


13,9  AMPS. 

DETERMINATION   OF  LOSSES 

KWln"  26,000 

KWrn"  127020X304,17(003    e'OI'IO'l  =26,836 

KWgN"  129,316  X  222,69  (003  1 7' 2e' 26-)  -  27,474 

LOSSES  TO  NEUTRAL 

CONDENSER                              ■•  "  600 

LOWEBINO  TRANSFORMERS 

(IRONLOSS)                                   •  "S'' 

(COPPER  LOSS  204, 1 7^x3, 1 86)  ».  ■  133 

RAISING  TRANSFORMERS 

(IRONLOSS)                                        .       "  236 

(COPPER  LOSS)  222.892x3,185           -  IBS 

HIGH  TENSION  LINE  27.474  -  (25.835  +  1 33)  ■  1 506 

TOTAL  LOSS  TO  NEUTRAL  -  2887 

EFFICIENCY 

25.966 
2  7,474 

EFFICIENCY.  (GENERATORS  TO  LOAD)  =    jf^ff 


=  94.52 
=  89  7r. 


KV-As- 
KV-Aoo" 


I  29.318  x222.69> 
1  29.3  I  8  X  1 47.09  ' 


'  86.393  PER  3  PHASE  CIRCUIT 
■■  67.064  PER  3  PHASE  CIRCUIT 


4.72  AMPS,- 
1 16.00  AMPS' 


ERROR  IN  RESULTS 

VOLTAGE  AT  SENDING  END  =-1,93% 

CURRENT  AT  SENDING  END  =+1.06% 

POWER  FACTOR  AT  SENDING  END  =  *-  1,02% 
KV-A  AT  SENDING  END  =-088% 

LOSS  IN  HIGH  TENSION  LINE  -•2.38% 

/  NOTE-  THESE  PERCENTAGES  OF  ERROR  ARE  BASED  UPON 
THTASSUMPTION  THAT  THE  COMPLETE  METHOD  PRO- 
DUCES 100%  VALUES. -THE  MINUS  SIGNS  SIGNIFY  RE- 
SULTS TOO  LOW:  THE  PLUS  SIGNS  RESULTS  TOO  HIGH. 


A  TYPICAL  220  KV  PROBLEM 


iji 


netizing  current  were  not  taken  into  account,  (as  it  also 

is  in  both  these  methods)  the  error  resulting  from  the 
use  of  the  approximate  method  would  be  considerably 
greater  than  the  above  values. 

The  simplified  graphical  approximate  solution  illus- 
trated by  Chart  XXIV  will  yield  results  sufficiently  ac- 
curate for  preliminary  work,  although  for  final  results 
it  should  be  supplemented  by  a  mathematical  solution 
and,  in  cases  of  very  long  lines,  a  complete  mathema- 
tical solution  might  be  desirable.  A  complete  solution 
as  given  by  Chart  XXIII  may  be  followed  as  a  guide  in 
such  cases. 

The  method  of  obtaining  the  auxiliary  constants 
corresponding  to  the  approximate  solution  is  given  be- 
low. The  linear  constants  of  the  circuit  including 
transformer  impedance  are  determined  as  follows: — 


A  = 


Sinh  01$      0.957   /o°27'oo" 

Tank  did      1.098    \i°oo'oo" 
=  0.8716  /x^^foo" 

a\  =  0.8713 

as  =  0.02206 
Sinh  0 


B  =  Z =  221.28    feo<'o9'23"  X  0.957  fa°vi'oo'' 

g  

=  211.76  /io'sG'zf'  ohms 
bi  =  34.561  ohms 
bi  =  208.92  ohms 

C=  Y =0.001211    I90   X0.957  MtYoo" 


Line 

Transformers 


Resistance 
(Ohms) 

...34.650 

...  3-i8S 


Reactance 
(Ohms) 

178.20 
39.82 


Total 37-835  218.02 

Dividing  these  total  values  by  225  we  obtain  the 

following  as  the  impedance  per  mile  of  the  combined 

circuit. 

r  =  o.i68i  ohms 
X  =  0.969  ohms 

TABLE   ZZZ— AUXILIARY    CONSTANTS    FOR   220   KV 
PROBLEM,  APPROXIMATE  SOLUTION 


Calculaled 


From 
Wilkinson  Chart 


From 
Keniielly  Chart 


ai  =  0.870783  =  100  % 

aj  =^0.02191 1  =  100% 
bi  =  34-5653  =  100% 
bs  =  2o8.83  =  ioo% 

Ci=r  -  0.000000  ^  100  % 
Ci  =  0.001158=  100% 


0.8921=102.44% 

0.868  =  99.68%  (corrected) 
0.0221  =  100.86% 
34-3  =  99-23% 
211.2^  101.14% 

-O.OOOOI  ^  III. II  % 

o.ooi  163  =  100.43  % 


0.8713  =  100.05  % 

0.02206=  100.68% 

34.561=99-99% 
208.92=  100.04% 

-0.000009  =:  100  % 

O.OOI  159  =  100.09% 


The  admittance  per  mile  is  assumed  the  same  as 
before  namely : — 


b  = 

=  5- 

.38  X  10- 

-''  mh< 

£r  = 

=  0 

'ilkinson 

's  Charts 

«i 

=  0.892 

and 

since 

rb 
ai 

=  0.904 

=  0.121 

bi 

=  34.3  ohms 

bi 

=  211. 2 

ohms 

and 

since 

rb-' 

=  4.865 

C\ 

=  —  0.000010 

1*2 

=  O.OOI 

63 

From  Dr.  Kennelly's  Charts 

z  =  37.835  +y  218.02 

=  221.28   /8o°09^23''^ 

Y  •=  o  +y  0.001 211 

=  0.001  211     190° 

ZY  =  0.26797  \i7o°09^23" 

9  =    VTY  =  0.5177     |85''04^4l'''' 

Sinh  e 
from  Chart  XIX  — ■_ —  =  0.957  /fo.45° 


d 


=  0.957  ^lYoo" 


TanhO 

from  Chart  XXI =  1.098  \i°oo'oo"* 


=  0.0011589    l90°27^oo^' 
Ci  =  —  0.000  009 
Ct  =>  0.001  159 

The  auxiliary  constants  as  obtained  graphically  and 
by  exact  mathematical  results  are  given  in  Table  ZZZ. 
The  same  remarks  in  regard  to  use  of  the  Kennelly 
charts  for  obtaining  the  auxiliary  constants  as  given 
under  the  complete  solution  also  apply  when  the  ap- 
proximate solution  is  used.  Wilkinson  chart  A,  if  used 
when  transformer  impedance  is  added  to  the  line  im- 
pedance, as  in  the  approximate  method,  requires  a  cor- 
rection to  constant  a^.  Constant  a^  as  read  from  this 
chart  will  be  correct  but  constant  a.^  as  read  from  the 
chart  will  be  too  high  for  the  following  rea- 
son. Constant  c^  accounts  for  the  rise  in 
voltage  along  the  line  at  zero  load  due  to  the 
charging  current  flowing  through  the  line 
inductance  adding  directly  to  the  sending 
end  voltage.  The  section  of  Wilkinson  chart 
A  applying  to  constant  a^  is  based  upon  dis 
tance  and  frequency  only,  so  that  values  read 
from  this  section  would  be  the  same  for  a  giv- 
en distance  and  frequency  regardless  of  whether  or  not 
transformer  impedance  is  included  with  the  line  con- 
stants. This  section  of  chart  A  therefore  takes  acount 
only  of  the  voltage  lowering  effect  of  the  charging  cur- 
rent flowing  through  the  line  inductance.  In  addition 
to  this,  it  flows  also  through  the  transformer  inductance, 
which  further  lowers  the  value  of  a^.  The  value  of  a, 
read  from  the  chart  must  therefore  be  reduced.  From 
the  chart,  Oj  =  0.892  volt  corresponding  to  a  voltage 
rise  of  0.108  volt  which  results  from  a  linear  conduct- 
ance reactance  of  178.02  ohms.  Actually  the  reactance 
of  the  circuit  including  lowering  transformers  is  218.02 
ohms  or  22.5  percent  greater.  Increasing  0.108  volt 
by  22.5  percent  we  get  0.132  volt  rise,  so  that  Oj  becomes 
i.ooo  —  0.132  =  0.868,  which  is  99.68  percent  of  the 
calculated  results. 

In  the  following  solutions  calculated  values  for  the 
auxiliary  constants  are  used  since  exact  results  are  re- 
quired for  the  purpose  of  comparing  the  results  with 
those  previously  obtained  by  the  complete  solution. 


♦This  was  interpolated  since  this  angle  lies  beyond  the  range 
of  this  chart. 


152 


A   TYPICAL  220  KV  PROBLEM 


EMERGENCY    LOAD COMPLETE    SOLUTION 

The  complete  solution  for  emergency  load  condi- 
tions shown  by  Chart  XXV  follows  the  same  construc- 
tion as  covered  by  Chart  XXIII  for  normal  load.  The 
difference  being  that  the  load  is  doubled  and  the  con- 
denser capacity  for  a  circuit  increased  nearly  four 
times.  Thus  to  force  double  the  amount  of  power 
through  the  line  and  transformer  impedance,  with  the 
same  voltage  drop,  it  is  necessary  in  this  case,  nearly 
to  quadruple  the  condenser  capacity  per  circuit.  Thus 
to  meet  the  emergency  condition  nearly  double  the  total 
condenser  capacity  will  be  required.  This  large  in- 
crease in  condenser  capacity  necessitated  drawing  the 
current  vectors  to  one  half  the  scale  used  for  current 
vectors  in  the  normal  load  diagram. 

EMERGENCY    LOAD APPROXIMATE    SOLUTION 

The    approximate    solution    for    emergency    load 
shown  by  Chart  XXVI  follows  the  same  construction  as 
in  Chart  XXIV  for  normal 
load  with   the    exception   of  5 
increased  load  and  condenser 
capacity. 

ZERO  LOAD — COMPLETE 
SOLUTION 

The  complete  solution 
for  zero  load  is  shown  by 
Chart  XXVII.  In  this  case  y 
the  load  is  made  up  of  a 
lagging  phase  modifier  load 
and  the  leakage  of  the  low- 
ering transformers.  The  same 
constructions  are  used  as  for 
the  other  complete  solutions. 

ZERO  LOAD APPROXIAIATE 

SOLUTION 

The  approximate  solu- 
tion for  zero  load  is  shown  by  Chart  XXVIII.  It  may 
be  seen  from  the  tabulated  errors  that  this  approximate 
method  produces  at  zero  load  larger  errors  than  the  cor- 
responding errors  for  loaded  conditions.  This  is  usual- 
ly of  little  importance,  however,  as  the  light  load  con- 
ditions are  generally  not  important. 

PHASE   MODIFIER    CURVES 

Frequently  the  normal  and  maximum  amount  of 
power  to  be  transmitted  is  known ;  that  is  the  transmis- 
sion line,  condensers  and  transformers  are  designed  for 
a  certain  maximum  load  and  it  is  of  little  importance 
what  condenser  capacity  would  be  required  for  other 
loads  or  for  various  sending  end  voltages.  At  other 
times,  especially  in  preliminary  surveys,  such  data  may 
be  very  necessary. 


In  Fig.  yo  are  plotted  curves*  showing  the  phase 
modifier  capacity  required  to  produce  certain  voltages 
at  the  sending  end  corresponding  to  various  receiving- 
end  loads  at  85  percent  power-factor  and  220  kv.  At 
85  percent  power- factor  and  220  kv  200000  kw  is  ap- 
proximately the  maximum  amount  of  power  which  may 
be  transmitted  through  the  lowering  transformers  and 
over  this  line  of  three  605  000  circ.  mil.  cables  if  the 
sending  end  voltage  is  not  permitted  to  exceed  230  lc\'. 
This  is  indicated  by  the  fact  that  the  curve  correspond- 
ing to  this  load  becomes  flat  when  it  reaches  the  230 
kv  horizontal  line.  To  deliver  this  maximum  load  at 
220  kv  through  the  impedance  of  this  line  will  require 
a  total  condenser  capacity  of  about  300000  kv-a.  The 
economic  capacity  of  the  line  is  reached  at  loads  very 
much  below  the  maximum  theoretical  limit  of  200  000 
kw. 

The  sending  end  voltages  corresponding  to  various 


0 

z 

~ 

r 

... 

z 

i- 

\ 

\ 

^ 

no  Z 

s 

^ 

V 

X 

^ 

N 

if,  z 

M 

^ 

s 

\ 

\ 

\ 

^ 

\ 

\ 

^ 

a 

fe"" 

\ 

\ 

\ 

\ 

^v 

'\ 

k 

\ 

^'( 

^. 

*«« 

HO   "* 

j^... 

\ 

\ 

K 

N 

^'t 

^ 

f^. 

V. 

^ 

0 

\ 

\ 

^ 

\ 

\ 

\ 

\\ 

^ 

^ 

^;^ 

'     "~ 

"-^ 

3 

Q   3*0 

z 

O«o 

\ 

\ 

\ 

"''l 

% 

-«^ 

<C'. 

Kl 

^ 

N. 

^ 

'~~' 

~- 

-^ 

Z 

) 

\ 

^ 

K 

H 

1** 

fS 

\, 

\ 

N 

-N 

"-^ 

\ 

™2 

u  *" 

% 

^ 

< 

% 

'ss 

f^ 

^ 

k. 

=^ 

s.,,^^ 

>.^ 

;;;:; 

■--, 

^ 

Hi 

> 

f<" 

% 

'•■ 

H 

f*t 

\ 

Xl 

N 

■~- 

L 

~ 

a  '"* 

N 

N 

x 

M 

o55~ 

30 

Doa 

' 5 ».ooo     ■    t 

lAOl* 

PHASE  MODIFIER    CAPACITY  REQUIRED-KV-A 
FIG.  70 — PHASE  MODIFIER  CAPACITY  REQUIRED  TO    MAINTAIN  CONSTANT  RECEIVER  VOLTAGE. 

These  curves  indicate  for  a  constant  load  power-factor  of  85  percent  lagging  and  con- 
stant load  voltage  of  220  kv,  the  amount  of  energy  which  may  be  delivered  to  the  load 
over  one  225  mile,  60  c}'cle,  three-phase  circuit  consisting  of  three  605  000  circ.  mil  alumi- 
num-steel conductors  corresponding  to  various  voltages  between  conductors  at  the  high-ten- 
sion side  of  the  raising  transformers.  The  values  by  which  these  curves  were  drawn  were 
determined  graphically.  For  230  kv  at  the  sending  end  the  maximum  amount  of  power  which 
can  be  transmitted  is  approximately  200  000  kw  and  to  force  this  amount  of  power  through 
the  line  impedance  will  require  approximately  300000  kv-a  capacity  in  phase  modifiers. 

capacities  of  phase  modifiers  in  parallel  with  different 
receiving  end  loads  for  drawing  curves  such  as  shown 
by  Fig.  70  are  most  readily  obtained  by  the  following 
graphical  procedure.  After  auxiliary  constants  A  and 
B  for  the  circuit  under  investigation  have  been  deter- 
mined (preferably  through  the  medium  of  both  the 
Wilkinson  and  Kennelly  charts)  a  tabulation  of  the  cur- 
rent to  neutral  corresponding  to  each  load  for  which 
curves  are  desired  is  made.  A  further  tabulation  of 
current  to  neutral  for  condensers  of  various  capacities 
is  made.  The  current  to  neutral  which  represents  the  loss 
in  the  various  condensers,  is  also  tabulated.    The  resist- 

*Such  curves  were  suggested  by  Mr.  F.  W.  Peek,  Jr.  in 
an  article  on  "Practical  Calculations  of  Long  Distance  Trans- 
mission Line  Characterictics"  in  the  General  Electrical  Review 
for  June,  1913,  p.  430. 


A   TYPICAL  220  KV  PROBLEM 


155 


CHART   XXV-220    KV   PROBLEM— EMERGENCY   LOAD 


RAISING  TRANSFORMEBS 
Zy^'l  6S26«il9  tlOHMS 


(COMPLETE  SOLUTION) 

(THIS  IS  DOUBLE  LOAD- 1 50.000  KW  AT  B5%  P  F  LAQOING) 


H10H  TtNSION  UNE 


2  "  M  •»  •  j  I  '8.2  OHMS 
Y  •    0        «j    0012  1 1    MHO 


U3WERIN0  TRANSfORMERS 
Z-]^"'  M36^jl0.9IOHM8 


HENDINQ  END 

EMERGENCY    LOAD 

PER  3  PHASE  PER  PHASE 


Tl    [   T    y   y    1    T    T   T   T    1   i    T   T    If- 


TO  NEUTRAL 


NEUTRAL 

TRANSFORMERS 

(FOUR  BANKS  IN  PARALLEL  AT  EACH  END  OF  THE  UNE) 


KV-Al-I'8.470 
KWi^- 1 60.000 
PF|_-85*LAO. 
El'  220,000 
||_-463.l 


KV-Aln=  68.824 
KWln"60  000 
PFln"86%LAG. 
Eln-12'020 

Iln»483.I 
00  CYCLES 

CONDENSERS 


R- 1^  =  1 .6926  OHMS  RESISTANCE 


19.91  OHMS  REACTANCE 


MAGNETIZING  CURRENT  -  27  8  AMPS  ATI  27  020  VOLTS 

IRON  LOSS       3.70  AMPS.  AT  I  27.020  VOLTS 

IRON  LOSS  •    470  KW  TO  NEUTRAL 

MAGNETIZING  CURRENT 
IRON  LOSS 


RECEIVING  END 

LINEAR  CONSTANTS 

Z  ■  34  86  «  i  I  78.2  OHMS 
V     0        *j  .001211    MHO 

'"    ™     AUXILIARY  CONSTANTS 

(a)  -  (a,  ♦  j  Sj)  .  .893,956  *  j  .020.234 
(B)  -  C"! 'j ''2)  "  82.198  ♦}  172.0*4  OHMS 

\(C)-(C|*J52)--  ooo.ooetj.ooi.i»«i»« 


FOUR  45.000  KV-A  SYNCHRONOUS 
CONDENSFRS  PER  3  PHASE  CIRCUIT  WHEN 
OPERATING  AT  FULL  LOAD  PROVIDE  MORE 
COMPENSATION  THAN  REQUIRED  FOR  MEETING 
THE  EMERGENCY    CONDITIONS-SINCE  BOTH 
3  PHASE  CIRCUITS  ON  THE  SAME  TOWERS 
WILL  BE  OPERATED  IN  PARALLEL  7  CONDENSERS 
MAY  BE    USED  FOR  THE  TWO  CIRCUITS.  THE  8TH 
CONDENSER  BEING  AVAILABLE  AS  A  SPARE.  UPON 
THIS  BASIS  THE  CONDENSER  DATA  PER  CIRCUIT  IS 
AS  FOLLOWS: 


KW, 


I, 


CALCULATION 
FOR  RECEIVING-END' 
CURRENT  AND  VOLTAGE 

l|_»  393  64 -j  243.94  AMPS.  TO  VECTOR   E| 


•     IS.63tJ4t3.0l 
T-      3.70 -i     27.80 


o-4l3.87  +  i  141  27 


COS  1 8"  60' 49' =  ,94638 
SIN     1 8- 60' 49' ".32304 

COS  1 4'  68'  0  r 

SIN     I4'68'0I'>. 26828 


-  437  32  / 1 8'  60'  49'  AMPS.  TO  VECTOR   E, 

-  437.32  /I4'68"0I  AMPS.  TO  VECTOR  OF  REFERENCE 

-  422.49  tj  112.94 

Ern  "   7  ( I  27.020  X  .94638  +  437.32  x  1.5926)^  +  { I  27.020  x  .32304  -  437.32  x  I  9.9  1)2 
-       126,152    /l4'680r  VOLTS  TO  VECTOR   | 

CALCULATION    FOR   HIGH   TENSION    CIRCUIT 

632  jp(A)- 376,42  ♦j  I  09.61 


Lg^ -1 06,047  ^j  78,876 

-I32.l64/3e'38'3r  VOLTS 


Is»37442tj266  69 

'463,40/ 34' 1 9' 44'  AMPERES 


DETERMINATION  OF  LOSSES 

KWln  .    60.000 

'^Wrn  -  1 26. 1 62  X  437.32  (COS  14'  68'  0  IT  -  82.874 

KWgN  -  132.184X463.40(008  2"  I8•47^  -  69.874 

'^Wqei^N-  133.662X460.47(003   9'2400T  -    80.871 

LOSSES  TO  NEUTRAL 

LOWERING  TRANSFORMERS  AND  CONDENSERS 

62.874-60.000  -    3874 

HIGH  TENSION  LINE  69.874 -62.874  ■    7000 

RAISING  TRANSFORMERS  80.97 1  -  69.874  -__7»7 

TOTAL  LOSS  TO  NEUTRAL  -10.671 

AS  A  PARTIAL  CHECK 

LOWERING  TRANSFORMERS 

(IRON  LOSS  3-7  X  127.02    ."  470 

(COPPER  LOS3437  32^x1,(936)         .  J04 

SYNCHRONOUS  CONDENSERS   ._ _- ..^     •  3100 

RAISING  TRANSFORMERS 

(IRON  LOSS  3,52  X  133,662)     -  470 

(COPPER  LOSS  463  4' X  1  6926)        -  327 

VIIQH  TENSION  LINE 

(VALUE  ABOVE  ASSUMED  AS  CORRECT  -  7000 
413.01  AMPS                                                              io«77 

i8  53AMPa       EFFICIENCY 

EfflOIENOY  (HIGH  TENSKW  LINE)         -   |f||J  •  88.3I« 
tFFlOlENCY,  (GENERATORS  TO  LOAD)-   |g  g°°  -  62.41* 


CALCULATION  FOR  GENERATOR  VOLTAGE  AND  CURRENT 


^OEN-N "  Vl  1 32. 1 84  x  .999 1 8  ♦  463.4  x  1 .6926)''  +  ( I  32. 1 84  x  .040359  t  463.4  x  1 9.9 1 )' 
=■  1 33.652  Iff  10' 23"  VOLTS  TO  VECTOR   Ig 
-  1 33.662  Ao'  30'  OT  VOLTS 

KV-As"  I  32. 1  84  X  463,40  x  3  -  I  70.789  PER  3  PHASE  CIRCUIT 
KV-AgEN"  I  33.662  X  460  47  x  3  -  |  64.490  PER  3  PHASE  CIRCUIT 
■^V-A  cc  *  '  32'  I  84  X  1 48.1 8  X  3  -     67.969  PER  3  PHASE  CIRCUIT 


4634  (.9942 -j.  1 0763) 

=  460.77 -j  48  76 


i  28  46  (LEAKAGE  OF  RAISING  TRANSFORMERS) 


'0EN-N"«4.=9-j76  2l 

■  480.47  \9'  24'  00'  AMPERES  TO  VECTOR  Eq£h 
-4S0>«7  /31'06  07'AMPtRe3 


I.S4 


A   TYPICAL  220  KV  PROBLEM 


CHART  XXVI— 220  KV  PROBLEM— EMERGENCY  LOAD 

(APPROXIMATE  SOLUTION) 

THIS  APPROXIMATE  SOLUTION  ASSUMES  THAT  THE  IMPEDANCE  OF  THE  LOWERING  TRANSFORMERS  MAY  BE  ADDED  TO  THE  LINE  IMPEDANCE  AND  TREATED  AS  THOUGH  IT  WERE  DISTRIBUTED  LINE  IM- 
PEDANCE--THIS  ASSUMPTION  SIMPLIFIES  THE  SOLUTION  AT  THE  EXPENSE  OF  ACCURACY  (SEE  LOWER  RIGHT  HAND  CORNER  OF  PAGE;  ALSO  TEXTl.— THE  SOLUTION  BELOW  IS  BASED  UPON  THE  VOLTAGE  BEINO 
HELD  CONSTANT  AT  THE  LOAD  SIDE  OF  THE  LOWERING  TRANSFORMERS  AND  AT  THE  HIGH  TENSION  SIDE  OF  THE  RAISING  TRANSFORMERS.--IF  THE  VOLTAGE  IS  TO  BE  HELD  CONSTANT  AT  THE  GENERATOR  BUS. 
THE  IMPEDANCE  OF  THE  RAISING  TRANSFORMERS  MUST  ALSO  BE  ADDED  TO  THAT  OF  THE  LINE-ALL  LOW  TENSION  VALUES  ARE  REFERRED  TO  THE  HIGH  TENSION  CIRCUIT. 


RAISING  TRANSFORMERS 
Z.^"l  6926  +  j  I9.ai  OHMS 

\l    (000000  I 


HIGH  TENSION  LINE 


34.66 +j  178.2  OHMS 
0        +j  .001211    MHO 


LOWERING  TRANSFORMERS 


SENDING  END 

EMERGENCY    LOAD 

PER  3  PHASE  PER  PHASE 


T  I  T  T  y  T  1  1  y  I  I    ' 


KV-Al=  I  76.470 

KWl"  1 60,000 

PFl=86%LAG. 


Il=463.  I 


TO  NEUTRAL 


KWln- 60.000 
PF|N  =  e6%LAG. 
=  127.020 
=  463.1 


NEUTRAL 

LINEAR  CONSTANTS 

Z  "37.S36+J2I8.02OHMS* 
Y=        0      +j. 001  21 1   MHO 
*  THIS  INCLUDES  IMPEDANCE  OF  LOWERING  TRANSFORMERS 

AUXILIARY  CONSTANTS 

(a)  =  (3|  +  j  ^2)  =  COSH  e  "  -870783  +  j  .02 1  9 1 


RECEIVING  END 

TRANSFORMERS 

(FOUR  BANKS  IN  PARALLEL  AT  EACH  END  OF  THE  LINE) 

H-jtf"  1 .6926  OHMS  RESISTANCE 

Xt^|=  I  9.9 1  OHMS  REACTANCE 

MAGNETIZING  CURRENT  -  27.8  AMPS.  AT  1 27,020  VOLTS 

IRON  LOSS =    3.70  AMPS.  AT  I  27.020  VOLTS 

IRON  LOSS .....  -   470  KW  TO  NEUTRAL 


60  CYCLES 

CONDENSERS 

FOUR  45,000  KVA  SYNCHRONOUS 
CONDENSERS  PER  3  PHASE  CIRCUIT     WHEN 
OPERATING  AT  FULL  LOAD  PROVIDE  MORE 
COMPENSATION  THAN  REQUIRED  FOR  MEETING 
THE  EMERGENCY  CONDITIONS-SINCE  BOTH 
3  PHASE  CIRCUITS  ON  THE  SAME  TOWERS 
WILL  BE  OPERATED  IN  PARALLEL  7  CONDENSERS 
MAY  BE  USED  FOR  THE  TWO  CIRCUITS.  THE  8TH 
CONDENSER  BEING  AVAILABLE  AS  A  SPARE,  UPON 
THIS  BASIS  THE  CONDENSER  DATA  PER  CIRCUIT 
IS  AS  FOLLOWS 

3  PHASE  TO  NEUTRAL 


470 

aos 


CONDENSERS 

LOWERING  TRANSFORMERS 

(IRON  LOSS!       ~ 

(COPPER  LOSS  437,32'  «  1. 6926) ... 
RAISING  TRANSFORMERS 

(IRON  LOSS! .„-■ -       470 

(COPPER  LOSS) 452.532  X  1. 5925         "        33^ 

HIGH  TENSION  LINE  60.248  -  (52.570  ♦  306)  ■     7373 

TOTAL  LOSS  TO  NEUTRAL  -  1 1 .044 

EFFICIENCY 

EFFICIENCY.  (HIGH  TENSION  LINE)     , 
EFFIOIENOV.  (CiENERATORS  TO  LOAD) 


62,876 
60.248 
60.000  , 
61.044 


87.76* 
6I.«I« 


%  ERROR  IN  RESULTS 


■  1 34.004  /44'  36'  07"  VOLTS 
KV- As   =  I  34.004  X  462,63  X  3  =  I  8  1 .922  PER  3  PHASE  CIRCUIT 
KV-Aco-  134.004X147.09X3=    69. 1  32  PER  3  PHASE  CIRCUIT 


VOLTAGE  AT  SENDING  END  "+1.39% 

CURRENT  AT  SENDING  END  "-0.20% 

POWER  FACTOR  AT  SENDING  END  -  -  0,66% 
KV.A  AT  SENDING  END  -+1.18% 

LOSS  IN  HIGH  TENSION  LINE  =  +  6,33% 

NOTE-  THESE  PERCENTAGES  OF  ERROR  ARE  BASED  UPON 
THE  ASSUMPTION  THAT  THE  COMPLETE  METHOD  PRO- 
^—  DUCES  100%  VALUES, -THE  MINUS  SIGNS  SIGNIFY  HE- 
\  SULTS  TOO  LOWi  THE  PLUS  SIGNS  RESULTS  TOO  HIGH. 
^ie.53AMP3 


Note  : — Linear  constant   Z,   as   used   in  this   chart,   incorrectly   includes  impedance  of  two  banks,  whereas  it  should  have  included 
four  banks  of  transformers.     This  error  will  not,  however,  materially  affect  the  result. 


A  TYPICAL  220  KV  PROBLEM 


155 


ance,  reactance,  iron  loss  and  magnetizing  currents  of 
the  transformer  banks  to  neutral  should  also  be  deter- 
mined for  all  capacity  transformer  banks  required. 
With  the  above  data  tabulated  any  draughtsman  can  be 
instructed  how  to  draw  vector  diagrams  of  the  circuit 
to  determine  the  sending  end  voltages  corresponding  to 


ficient  to  locate  the  curve,  although  more  points  were 
calculated  for  drawing  the  curves  of  Fig.  70.  This 
method  of  obtaining  condenser  capacities  correspond- 
ing to  sending  end  voltages  is  a  cut  and  try  method. 
Il  has  one  important  advantage  in  its  favor.  That  is, 
the  results  check  each  other,  so  that  an  error  in  one 


CHART   XXVM— 220    KV   PROBLEM—ZERO    LOAD 

(COMPLETE  SOLUTION; 
(THIS  OORRESPONOS  TO  NORMAL  LOW  OONNECTJONSI 

*T  ZCRO  1.0*0.  WITH  230.000  VOITS  MAINTAINED  BETWCCN  CONDUCTORS  (132.T8T  VOLTS  TO  NtUTRALl.  AT  THE  MIGM  TENSION  SIDE  OF  THE  RAISING  TRANSfOHMERS  THE  VOtTAGC  AT  THE  HIGH  TCI*. 
SION  SIDE  Of  THE  LOWERING  THANSEORMERS  [NEGLECTING  THE  EFFECT  OF  THE  LAGGING  MAGNETIZING  CURRENT  OF  THE  LOWERING  TRANSFORMERSi  WILL  RISE  TO  230.000  DIVIDED  ST  Ai- 230,000  DIVIDED  ST 
.89416  ■  25T.219  VOLTS  BETWEEN  CONDUCTORS  HAB.SIO  VOLTS  TO  NEUTRAL).  ACTUALLY  THE  GREATLY  INCREASED  LAGGING  MAGNETIZING  CURRENT  OF  THE  LOWERING  TRANSFORMERS  WHEN  EXCITED  BT  AB- 
NORMALLY HIGH  VOLTAGE  WILL  NOT  PERMIT  OF  THE  RECEIVING  END  VOLTAGE  REACHING  SUCH  A  HIGH  VOLTAGE  UNLESS  THE  GENERATOR  VOLTAGE  RAISES  MOMENTARILY  TO  A  VALUE  GREATLY  IN  EXCESS  Of 
230,000  VOLTS.  IF  HOWEVER  THE  LOWERING  TRANSFORMERS  ARE  DISCONNECTED  FROM  THE  CIRCUIT,  THE  INCREASED  LEADING  CHARGING  CURRENT  OF  THE  LINE,  REACTING  UPON  THE  GENERATOR  FIELDS, 
CCVleiNED   WITH    A    MOMENTARY   OVER    SPEED   OF    THE    GENERATORS    MAY  CAUSE   THE  RECEIVING  END  VOLTAGE  TO  GREATLY  EXCEED  THE  ABOVE  VALUE. 

IN  ORDER  TO  HOLD  THE  VOLTAGE  AT  THE  RECEIVING  END  CONSTANT  AT  220,000  VOLTS  BETWEEN  CONDUCTORS  1127.020  VOLTS  TO  NEUTRALI  AT  ZERO  LOAD  IT  WILL  BE  NECESSARY  TO  PLACE  AN  ARTL 
FICIAL  LAGGING  LOAD  AT  THE  LOAD  END  OF  THE  LINE--THIS  IS  ACCOMPLISHED  BY  OPERATING  ONE  OF  THE  SYNCHRONOUS  CONDENSERS  WITH  ITS  FIELDS  UNDER  EXCITED— BY  CONSTRUCTING  SEVERAL  VECTOR 
DIAGRAMS  FOR  THIS  CIRCUIT  EACH  BASED  UPON  DIFFERENT  VALUES  OF  REACTOR  LOAD.  A  CURVE  MAY  BE  DRAWN  BY  PLOTTING  THE  REACTOR  LOADS  AGAINST  THE  CORRESPONDING  BCNOING  END  VOLTAOCS.— 
FRON  THIS  CURVE  THE  REACTOR  CAPACITY  CORRESPONDING  TO  23O.0O0  VOLTS  BETWEEN  CONDUCTORS  AT  THE  SENDING  END  WILL  BE  SEEN  TO  BE  APPROXIMATELY  30.000  RV-A 


LEAKAGE  OF 

RAISING 
TRANSFORMERS 


CONDENSER 

(ONE  RE(3UIRED) 
3  PHASE  TO  NEUTRAL 

KV-A(;=  30.000  KV-AoN  =  io.ooo 

Ec=220.000  EcM=U7.O20 


0=78,73 


KWc 


KW, 


'ON- 


0-LOSS-N  '  508 
'o-LOSS-N'=*00 


LINEAR  CONSTANTS 

Z  =  34,66  tj  178,2  OHMS 

Y=    0        +j  .001211    MHO  gig        2g 

AUXILIARY  CONSTANTS  ?>f,     M 

(a)  -  (8l +j  82) -OOSH  8  =  . 893.966 +j. 020.234"   .BS4I8    /l'(7'47'"  j         _,^ 
(B).(b|*jbj)  =  25Mil=yisiNH6.32.l98tjl72.094OHM8     |S        |S 


,=  -l.04+jl6269  AMPERES 
'q£N-N   =57.1  I  /si'Se'CA"  AMPERES 

PFqen"'*°*i-"°"*° 


(C)-(c,*jOj).y§^. 

WHERE  6"    fZY 


=  siNH8- -.ooo.ooe+j.ooi,ie8  MHO  5s 


0SO" 


84-  44'  3  I  •  PFso  -  9  '  **  LEADING 


/^gen-no"  "°-'°°    /  0' 30'  1 6"  VOLTS 


^-(vfclOHOmt^ERENOE)       VE,^- 1 27.020  VOLTa  "^EgNo  "  ' 

CALCULATION    FOR 
RECEIVING-END  CURRENT  AND  VOLTAGE 

Ico"  *00-i'8.e3  COS,  86' 22' 67- -.063096  COS.  Be"  21' 27- -  .08363 
|.^P=  1.86-j  13,90  SIN.  86' 22' 67"  =9980  I  SIN.  86"  21' 27' -  .99798 
|p|-l=    6  86- j  92.63  AMPERES 

=  92.71  \86'  22'  67"  AMPERES  TO  VECTOR  E^no 


i.»7«/0'ir36'  VOLTS 


LEAKAGE  OF 

LOWERING 

TRANSFORMERS 


Ion  -  92.7  I    \86'2|-27-  AMPERES  TO  RECTOR  OF  REFERENCE 
-    6.89 -j  92.62  AMPERES 

Erno"  7  "  ''•°^°  "  °"°^5  +  92,7 1  k  3. 1  86)^  +  ( I  27.020  «  .9980 1  +  92.7 1  «  39.62)^ 
•      1 30.724  LSS11112i"  VOLTS  TO  VECTOR  Ifio 

CALCULATION    FOR   HIGH   TENSION    CIRCUIT 


DETERMINATION   OF  LOSSES 
KWlno      -0 

I^Wrno  -  130.724  k  92.7 1  ((XJ8  6«"  8  r  27n  •  7T0 
^'^ma  >l32.974K703a((X»W'44'3n  -  MT 
KWqe,^.,^- 1 30  206  K67.il  (008  8 r 26' 4**)  -  MOt 

LOSSES  TO  NEUTRAL 

tOWERINO  TRANSFORMERS  AND  CONDENSER  -  770 

tW3H  TENSION  LINE  667-770                            ,  "  87 

RAiaiNQ  TRANSFORMERS  I  1 08-867                  -  861 

'TOTAL  LOSSjrO  NEUTRAL  -  1 108 

AS  A  PARTIAL  CHECK 


-RN0(A)=  I  16.861  +j2e46 

1ro(B)  =   ie.iii-ii966 

EsNO'  1 32.972  tj  680 


Iro(A)=      7.13  -  j    8269 

^RN0(C)=  -  1.04  *i  162  69 

|gn=      8.09  +  j    70.10 


I  C(}S. -.091642 


-  132.974/ 0' I  7' 36'  vni  T.-i  -  70.36  /  86' 02' 06"  AMPERES 

CALCULATION  FOR  GENERATOR  VOLTAGE  AND  CURRENT 

^QEN-NO-   I  (I  32,974  X  09  I  642  4.  70,36  K  3, 1  SSl''  *  ( I  32,974  x  .99679  -  70.36  x  39.82)' 


LOWERINO  TRANSFORMERS 

(IRON  LOSS  1,86X127.02 ...   - 

93S 

(COPPER  LOSS  82,71 '»1.H6) - 

SYNCHRONOUS  CONDENSER 

RAISINO  TRANSFORMERS 

(IRON  LOSS  1,80  X  130,308)           .  «. - 

(COPPER  LOSS  70  3«'x3.||6) - 

HIGH  TENSION  LINE 

(VALUE  ABOVE  ASSUMED  AS  CORRECT  -  _ 

IT 
•0( 

iis 

16 

87 

1108 

EFFICIENCY 

EFFICIENCY,  (HIOH  TENSION  LINE)         "517" 

ef>.8(» 

1  30.206  /84'3r  60'  VOLTS  TO  VECTOR 
1 30.206    /  0*30'  16'  VOLTS 


70.38  (.0963 1 6 -fj. 99646) 
=    6,71  +j    70,04 

-  1,80 -j     13,67    (LEAKAGE  OF  RAISINO  TRANSFORMERS) 
'qEN-N-    8  61   *j    66,47 

-67,11  /8r26'48'  AMPERES  TO  VECTOR  Eqe^-no 

-  67, 1  I  /8I'66'04'  AMPERES 


.  (COS.  =  .096316 
I  SIN.    -.99646 


C(3S.SI'26'48'-    14902 

KV-Aso°  '  32  974  X  70.36  x  3  -  28.068  PER  3  PHASE  CIRCUIT 

KV-Aqen-o"  '  30.206  X  67. 1  I  X  3  -  22.308  PER  3  PHASE  CIRCUIT 

KV-Aco"  132.974(001  168  xU8,6IO)x  3  =  69.197  PER  3  PHASE  CIRCUIT  * 
ABASED  UPON  H.T.  LINE  BEING  OPEN  AT  RECEIVING  END. 


the  various  receiving  end  loads  and  different  phase  mod- 
ifier capacities. 

The  graphical  method  used  in  determining  the 
values  to  plot  the  curves  of  Fig.  70,  is  illustrated  by  Fig. 
71.  Three  solutions  are  illustrated,  two  with  condens- 
ers of  different  size  and  one  without  condensers. 
Three  such  solutions  for  each  load  will  usually  be  suf- 


of  the  graphical  constructions  corresponding  to  a  given 
load  will  be  detected,  since  the  point  will  not  lay  in  the 
curve  and  an  error  in  a  curve  corresponding  to  a  given 
load  will  be  detected  by  the  curves  of  Fig.  72. 

CAPACITY  OF  PHASE  MODIFIERS 

The  curves  of  Fig.  70  show  that,  for  a  constant 
delivered  load,  power-factor  and  voltage,  the  leading 


156 


A   TYPICAL  220  KV  PROBLEM 


capacity  of  phase  modifiers  required  goes  down  as  the 
Hne  drop  increases.  For  instance  75  000  kw  at  85  per- 
cent power-factor  and  220  kv  can  be  delivered  over  this 
line  with  230  kv  sending  end  voltage,  if  43  000  kv-a 
condenser  capacity  is  placed  in  parallel  with  the  load. 
If,  however,  a  line  drop  of  20  kv  is  selected  in  place 
of  10  kv,  the  sending  end  voltage  will  be  240  kv  and 
the  corresponding  condenser  load  will  be  reduced  to 
approximately  30000  kv-a.  On  the  other  hand  this 
increased  line  drop  will  necessitate  a  greater  capacity 


The  dotted  line  in  Fig.  70  is  simply  the  zero  load  line 
thrown  over  to  the  leading  load  side  to  facilitate  studv 
in  phase  modifier  capacity.  For  instance,  projection 
from  the  points  where  the  dotted  line  intersects  a  load 
curve  will  give  the  minimum  capacity  of  phase  modifier 
on  the  bottom  scale  and  the  corresponding  sending  end 
voltage  on  the  vertical  scale  to  the  left.  Thus  with  a 
load  of  75  000  kw,  intersection  of  the  dotted  line  with 
this  load  curve  indicates  that  33  000  kv-a  phase  modi- 
fier capacity  will  be  required  both  at  this  load  and  at  zero 


CHART   XXVMI— 220   KV   PROBLEM— ZERO   LOAD 

(APPROXIMATE  SOLUTION) 

(THIS  CORRESPONDS  TO  THE  NORMAL  LOAD  CONNECTIONS) 

THIS  APPROXIMATE  SOLUTION  ASSUMES  THAT  THE  IMPEDANCE  OF  THE  LOWERING  TRANSFORMERS  MAY  8E  ADDED  TO  THE  LINE  IMPEDANCE  AND  TREATED  AS  THOUGH  IT  WERE  DISTRIBUTED  UNE  IM. 
PCDANCE-THIS  ASSUMPTION  SIMPLIFIES  THE  SOLUTION  AT  THE  EXPENSE  OF  ACCURACT  (SEE  LOWER  RIGHT  HAND  CORNER  OF  PAGE:  ALSO  TEXTI.-THE  SOLUTION  BELOW  IS  MSEO  UPON  THE  VOLTAGE  BEINC 
MELD  CONSTANT  AT  THE  LOAD  SIDE  OF  THE  LOWERING  TRANSFORMERS  AND  AT  THE  HIGH  TENSION  SIDE  OF  THE  RAISING  THANSFORMERS.-IF  THE  VOLTAGE  IS  TO  BE  HELD  CONSTANT  AT  THE  GENERATOR  BUS 
THE  IMPEDANCE  OF  THE  RAISING  TRANSFORMERS  MUST  ALSO  BE  ADDED  TO  THAT  OF  THE  LINE-ALL  LOW  TENSION  VALUES  ARE  REFERRED  TO  THE  HIGH  TENSION  CIRCUIT  ■  Ht  glhehator  BUS. 

AT  ZERO  LOAD.  WITH  ?3O.000  VOLTS  MAINTAINED  BETWEEN  CONDUCTORS  (I32.T87  VOLTS  TO  NEUTRAL).  AT  THE  HIGH  TENSION  SIDE  OF  THE  RAISING  TRANSFORMERS  THE  VOLTAGE  AT  THE  HIGH  TEN- 
8.0N  SIDE  OF  THE  LOWERING  TRANSFORMERS  .NEGLECTING  THE  EFFECT  OF  THE  LAGGING  MAGNETIZING  CURRENT  OF  THE  LOWERING  TRANSFORMERS? vlrLL  RISE  TO  230  OOOOIvVoEDByTa"/^^^  DmSED  BY 

■  7I0S  -ZSA.OAS  VOLTS  BETWEEN  CONDUCTORS  152  »5tVOLTS  TO  NtUTRALi.  ACTUALLY  THE  GREATLY  INCREASED  LAGGING  MAGNETIZING  CURRENT  OF  THE  LOWErTnG  TRANSFORMERS  WHEN  EXcfTED  BY  AB- 
NORMALLY HIGH  VOLTAGE  WILL  NOT  PERMIT  OF  THE  RECEIVING  END  VOLTAGE  REACHING  SUCH  A  HIGH  VOLTAGE  UNLESS  THE  GENERATOR  VOLTAGE  RAISES  mSmentARIlV  TO  A  VaIuE  GREATLY  IN  EXCESS  OF 
i^^^.Z:'^'-!^^".  'i?,'lV'«S".J?'„l;?S'=="^;?Jj;''#i?".!t=„"Aj;S.°J^£?~~,'^"^''  =  '>/''°"  ■'"■^  CRCUIT.THE  increased  leading  charging  CURRENT  OF  THE  LIiSe.  SeACtIngui^NtSe  GENERATOR  F|"dS, 
COMBINED  WITH   A   MOMENTARY    OVER    SPEED  OF  THE    GENERATORS    MAY  CAUSE  THE  RECEIVING  END  VOLTAGE  TO  GREATLY  EXCEED  THE  ABOVE  VALUE 

IN  ORDER  TO  HOLD  THE  VOLTAGE  AT  THE  RECEIVING  END  CONSTANT  AT  220.000  VOLTS  BETWEEN  CONDUCTORS  (127,020  VOLTS  TO  NEUTRAL!  AT  ZERO  LOAD  IT  WILL  BE  NECESSARY  TO  PLACE  AN  ARTI. 
riCIAL  LAGGING  LOAD  AT  THE  LOAD  END  OF  THE  LINE-THIS  IS  ACCOMPLISHED  BY  OPERATING  ONE  OF  THE  SYNCHRONOUS  CONDENSERS  WITH  ITS  FIELDS  UNDER  EXCITeS  .-BY  CONSTRlJCTING  SEVERAL  VECTOR 
DIAGRAMS  FOR  THIS  CIRCUIT  EACH  BASED  UPON  DIFFERENT  VALUES  OF  REACTOR  LOAD,  A  CURVE   MAY  BE  DRAWN  BY  >Co-fTING  THE  REACTOR  LOADS  AcjArNST  THE  COR^  END  "oLTAMB  ." 

FHON  THIS  CURVE  THE  REACTOR  CAPACITY  CORRESPONDING  TO  230.000  VOLTS  BETWEEN  CONDUCTORS  AT  THE  SENDING  END  WiLl  BE  SEEN  TO  BE  APPROXIMATELY  30  OOO  RV-A  ^  *     "*  VOLTAWB... 


CONDENSER 

(ONE  REQUIRED) 


'c-LOSS'*<"' 


to  neutral 
KV-A(;n=i<"""' 


KW, 


ON'"" 

c-loss-n"'"' 
'c-Loss-Nr*"" 


LINEAR  CONSTANTS 

Z  '  37.S36  *j7\  8.03  OHMS  -k 
Y  -       0      +i.OOI2ll  MHO 
1l  THIS  INCLUDES  IMPEDANCE  OF  LOWERINO  TRANSFORMERS 

AUXILIARY  CONSTANTS 

(A)  =  (ai*iaa)='  cosh e  =  . 670783 -tj. 021 91 1»  87108  /r  ae'  7%- 


SINH6 
0 


/•C-SNO 


(B).(bcjbj).z 
{C)-(0i*iCs).y5!Mie 

WHERE  e  -   ^ZY 


130.134  •  /0"!2II7'  VOLTS 


■ft 


SINH  e  -  34.8883  +  j  208.83  OHMS 
8INH  a  "  -  .000.008  «j  .00 1 1 88  MHO. 


CALCULATION   FOR 


MVECTOR  OF  REFERENCE) 

DETERMINATION  OF  LOSSES 
KW,«o-o 


RECEIVING-END  CURRENT 


RNO' 


I27.030XS2.7I  (COS  Be' 22' 67-|-743 


.0«°    pfno 


'CO 

'to 
'ro 


=    4.00 -j  78.83 
•    186-113.80 


6.88  -j  92.63  AMPERES 
92.7 1  \88'2287-  AMPERES 


CALCULATION    FOR   HIGH   TENSION   CIRCUIT 


LEAKAGE  OF 

LOWERING 
TRANSFORMERS 


ErNO  (A)=  I  1 0,607 +j  2783 

lRf%(B)°     I9S26-J    1977 

ESNO'  I30.l32*j    806 

=  130.134  /O'  21'  17'  VOLTS 


!ro(A)- 

Erno^C)= 

'so" 


■  ),l4+j  147,09 
6.98 +j  86.66 
88.92  /  84- 62  23'  AMPERES 


KV-A  so  -  I  30. 1  34  X    68.92  X  3  -  26. 1  26  PER  3  PHASE  CIRCUIT 


KV-A  CO  =  I  30. 1  34  (.00 II  68  X  I  62,46  I )  «  3  =  69,5  I  6  PER  3  PHASE  CIRCUIT  •*■ 
if  BASED  upon  H.T.  LINE  BEINQ  OPEN  AT  RECEIVING  END. 


KW, 

KWjNo"  1 30. 1 34  «  66  82  (008  84"3r08")-B3l 

LOSSES  TO  NEUTRAL 

CONDENSER                            -  808 

LOWERINQ  TRANSFORMERS 

(IR0NL(^8> -»  836 

(COPPER  LOSS  92.7l'»3.ie6) -  ■  «7 

RAISING  TRANSFORMERS 

(IR0NL0S3)        > ,..-  218 

(COPPER  LOSS  66  82^x3.186)  .»  14 

HIGH  TENSION  LINE  83 1 -(743  +  27)       " 6^ 

TOTAL  LOSS  TO  NEUTRAL  -  1080 

EFFICIENCY 

EFFICIENCY.  (HIOH  TENSION  LINE)       .-   ^-92,88% 

%  ERROR  IN  RESULTS 

VOLTAGE  AT  SENDING  END  "  -     2.13% 

CURRENT  AT  SENDING  END  "-     4.86% 

POWER  FACTOR  AT  SENDING  END  -  ♦     4,24% 

KV-A  AT  SENDING  END    ---     6,93* 

LOSS  IN  HIGH  TENSION  LINE  ..--2088% 

NOTE-  THESE  PERCENTAGES  OF  ERROR  ARE  BASED  UPON 
THE  ASSUMPTION  THAT  THE  COMPLETE  METHOD  PRO- 
DUCES 100%  VALUES.-THE  MINUS  SIGNS  SIGNIFY  RE- 
SULTS TOO  LOV*(  THE  PLUS  SIGNS  RESULTS  TOO  HIGH. 


at  zero  load  in  order  to  maintain  240  kw  constant  at 
the  sending  end.  Thus  with  230  kv  at  the  sending  end, 
about  30  000  kv-a  reactor  load  will  be  required  at  zero 
load,  whereas  with  240  kv  at  the  sending  end,  about 
40  000  kv-a  reactor  load  will  be  required  at  zero  load. 

Obviously  the  smallest  phase  modifier  capacity 
possible  to  maintain  regulation  is  one  in  which  full  ca- 
pacity leading  will  be  required  under  maximum  load 
and  full  capacity  lagging  under  zero  load.  At  half  load 
such  a  phase  modifier  would  operate  at  near  zero  kv-a. 


load  and  that  the  corresponding  sending  end  voltage  will 
be  approximately  236  kv.  At  100  000  kw  load,  nearly 
50000  kv-a  phase  modifier  capacity  will  be  required, 
and  the  corresponding  sending  end  voltage  would  be 
250  kv. 

As  previously  stated,  phase  modifiers  which  may  be 
operated  at  rated  load  both  lagging  and  leading  are  spe- 
cial, and  cost  more  than  standard  phase  modifiers.  On 
account  of  unstable  operation  due  to  weakened  field, 
standard  condensers  usually  cannot  be  operated  at  lag- 


A  TYPICAL  220  KV  PROBLEM 


1S7 


ging  loads  above  approximately  70  percent  of  their  full 
load  leading  rating.  To  deliver  75  000  kv-a  at  85  percent 
power-factor  requires  approximately  42000  kv-a  in 
phase  modifier  capacity  with  230  kv  at  the  sending  end. 
To  maintain  the  sending  end  voltage  of  230  kv  at  zero 


liMM 

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BM         U 

which  determines  the  total  capacity  of  phase  modifiers, 
for  the  220  kv  problem.  For  instance  at  normal  load, 
43  000  kv-a  in  capacity  is  required,  whereas  for  the 
double  or  emergency  load  157000  kv-a  capacity 
(nearly  four  times)  is  required  This  large  increase 
is  due  to  the  fact  that  the  line  charging  cur- 
rent (which  tends  to  reduce  phase  modifier 
capacity  under  load)  has  not  changed,  and 
that  the  line  impedance  volts  has  become 
twice  as  much,  making  it  necessarj'  to  turn 
the  line  impedance  triangle  through  a  large 
angle  in  the  counter-clockwise  direction  in 
order  that  the  sending  end  voltage  be  not  in- 


creased. 


PHASE  MODIFIER  CAPACITTf  REOUrRED-KV-A  -LEADING 


FIG.    72 — PHASE    MODIFIER   CAPACITY   REOUIRICD  FOR  THE  VARIOUS  LOADS 

These  curves  are  plotted  from  values  read  from  the  curves  of  Fig. 
70  and  are  on  the  basis  of  a  constant  load  voltage  of  220  kv. 


load  requires  approximately  30000 
kv-a  lagging.  This  is  70  percent  of 
the  capacity  leading,  thus  permitting 
of  employing  a  standard  43  000  kv-a 
condenser.  To  provide  margin  a 
45  000  kv-a  standard  condenser  might 
be  selected  for  this  normal  load  con- 
dition. 

Under  emergency  conditions 
(that  is,  double  or  150000  kw  load 
at  85  percent  power-factor)  157000 
kv-a  phase  modifier  capacity  will  be 
required  if  230  kv  is  not  to  be  ex- 
ceeded at  the  sending  end.  If  the 
generator  can  be  operated  during  the 
emergency  condition  at  increased 
voltage  of,  for  instance,  240  kv,  the 
phase  modifier  capacity  could  be  re- 
duced to  approximately  140000 
kv-a.  However,  too  much  liberty  in 
variation  of  generator  operating 
voltage  should  not  be  taken.  If  the 
voltage  is  held  constant  at  the  high- 
voltage  side  of  the  raising  transform- 
ers, the  generator  operating  voltage 
will  have  to  be  varied  to  compensate 
for  the  regulation  of  the  sending  end 
transformers,  and  to  provide  a  still 
greater  range  in  generator  operating 
voltage  might  impose  a  hardship  on 
the  generator  designers.  The  volt- 
age drop  through  the  transformers 
is  small  under  load  conditions,  since 
the  power-factor  will  be  near  unity, 
but  under  zero  load  condition  the 
drop  will  be  considerable,  due  to  the 
low  power-factor,  especially  if  a 
large  phase  modifier  load  is  re- 
quired at  zero  load.  It  will  be  seen 
that    it    is    the   emergency   condition 


CONSTANTS 


E^-  VOLTUC  AT  LOAD  HElA  OOMSTAnT  at  3,0  oOO  VOLTS 
KWt"''00OHW*Ti«PFOON»T*NT   80TMAT  1^   •  III  \]|-4r*M^ 

(A)  -  ••**;  0I03.(S|.j«j) 

(B)  -3J  J'ilH0MM8.(b,  .jD,) 

(C)  -  l»*OT  ItEQUmED  MEHEt 
W'TM  M  OCC  KV  »  COWDENaEI  C*y*OITV 


TRANaroAMER  nESlSTANCt  -  3  <  ■  OHMS  TO  MFUTWAL 
TTUNHFOAMER  REACTANCE  •  3*  •]  OHMS  TQ  NEUTRAl 
TRANSrOAMER  LEAKAGE  •  I  »  -  j  1  3  »  AMPS  TO  NEUTRM. 


FIG.  71 — GRAPHIC  METHOD  FOR  DETERMINING  THE 
VOLTAGE  AT  THE   SENDING   END. 

Corresponding  to  different  condenser 
loads  in  parallel  with  a  constant  power  load  of 
75  000  kw  at  85  percent  power- factor  and  220 
kv.  The  results  as  plotted  in  Fig.  70  were 
obtained  by  similar  constructions. 


LINE  1I2.S  MILES  LONG 
(A)-.T..,«.i 


LINE  22S  MILES  LONG 
(A).  •••.,«•» 
(B)-...-, 


LINE  337.6  MILES  LONG 


E-" 

FIG.  Ti — VECTOR  DIAGRAMS  SHOW- 
ING THE  EFFECT  OF  THE  LENGTH 
OF  THE  LINE  ON  THE  PHASE  MODI- 
FIER CAPACITY   REQUIRED 

The  diagrams  represent  a 
three-phase,  60  cycle  circuit,  con- 
sisting of  three  605  000  circ  mil 
aluminum  steel  reinforced  con- 
ductors, when  delivering  75  000 
kw  at  85  percent  lagging  power- 
factor  at  a  load  voltage  of  220  kv 
with  a  sending  end  voltage  of  230 
kv. 


158 


A  TYPICAL  220  KV  PROBLEM 


The  zero  load  curve  on  Fig.  70  is  drawn  for  the 
normal  load  connection;  that  is,  for  two  50000  kv-a 
transformer  banks  in  parallel.  For  the  emergency  load 
four  transformer  banks  in  parallel  will  be  required. 
The  result  of  the  increased  magnetizing  current  con- 
sumed by  four  in  place  of  two  transformer  banks  will 
be  to  reduce  the  capacity  of  phase  modifiers  required 
under  zero  load.  A  second  zero  load  line  could  be 
added,  covering  four  transformer  banks.  Such  a  line 
would  lie  directly  above  the  one  for  two  transformer 
banks  but  would  not  materially  affect  the  results.  For 
load  conditions  of  100  000  kw  at  85  percent  power-fac- 


LINC  112.6  MILES  LON& 


PHASE  MOOinER  CAPACITY  RESUIREO-KVA 
FIG.   74— CURVE  SHOWING  THE  RELATION  BET\VEEN  PHASE  MODIFIER 
CAPACITY    AND    SENDING    END   VOLTAGE 

For  various  receiving  end  loads  of  85  percent  lagging 
power-factor  and  a  constant  load  voltage  of  220  ky.  These 
carves  apply  to  a  three  phase,  60  cycle  circuit  consisting  of 
three  605  000  circ.  mil  aluminum  steel  conductors.  The  vector 
construction  of  these  four  lines  is  shown  in  Fig.  73. 

tor  and  above,  the  points  for  the  curves  were  determined 
on  the  basis  of  four  transformer  banks. 

In  the  above  it  was  assumed  that  the  power-factoi 
of  the  load  would  be  85  percent  lagging.  A  long  line 
such  as  this  would  probably  feed  into  an  extended  dis- 
tribution net  work,  having  numerous  load  centers.  At 
these  load  centers  synchronous  condensers  would  pro- 
bably be  located  for  the  purpose  of  holding  the  voltage 
constant.  This  would  necessitate  operating  the  con- 
denser leading  at  heavy  loads  thus  raising  the  power- 
factor  of  the  entire  system  under  load,  and  in  effect 
reducing  the  capacity  of  phase  modifiers  required  for 
voltage  control  at  the  receiving  end  of  the  line.  This 
point  should  be  investigated  where  a  long  line  such  as 


this  feeds  a  net  work  on  which  condensers  are  required 
for  voltage  control. 

It  may  be  desired  to  investigate  the  effect  of  line 
charging  current  on  phase  modifier  capacity  for  lines  of 
different  lengths.  For  this  purpose  the  vector  dia- 
grams Fig.  73,  and  the  phase  modifier  curves,  Fig.  74, 
were  prepared.  These  vector  diagrams  and  curves  are 
based  upon  a  constant  load  of  75  000  kw  at  85  percent 
power-factor  delivered  at  220  kv  and  a  line  drop  of  10 
kv.  In  other  words  the  only  variable  for  the  four 
different  lines  is  the  length  and  this  varies  in  equal  in- 
crements. 

The  vector  diagrams  of  Fig.  73  show  the  influence 
of  line  charging  current  upon  condenser  capacity.  As 
the  length  of  the  line  increases,  the  influence  of  the  in- 




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FIG.  75 — CURVES  SHOWING  THE  VOLTAGE  ON  EACIt 
SIDE     OF     THE     RAISING     TRANSFORMERS 

Corresponding  to  condenser  loads  of  vari- 
ous capacities  in  parallel  with  a  constant  load 
of  75  000  kw  at  85  percent  power  factor  lagg- 
ing and  220  kv.  The  vertical  distance  between 
the  two  voltage  lines  is  the  voltage  drop  or 
voltage  rise  through  the  raising  transformers. 
For  condenser  loads  up  to  15000  kv-a  there  is 
a  drop  in  voltage  through  the  raising  transfor- 
mers. For  condenser  loads  above  15000  ky-a 
there  is  a  rise  in  voltage  through  the  raising 
transformers. 

creased  line  charging  current  is  toward  a  reduction  in 
condenser  capacity;  that  is  the  line  itself  furnishes  a 
large  part  of  the  leading  current  necessary  to  maintain 
the  proper  line  voltage  drop.  If  this  line  were  longer 
than  450  miles,  the  line  charging  current  at  a  certain 
length  would  be  sufficient  in  itself  to  maintain  the  de- 
sired voltages  at  the  two  ends  of  the  line  without  the 
aid  of  condensers.  In  such  a  case,  however,  a  large 
reactor  capacity  would  be  required  at  zero  and  low 
loads  to  hold  the  receiving  end  voltage  at  a  constant 
value. 

The  reason  that  a  short  line  may  necessitate  more 
condenser  capacities  for  voltage  control  than  a  long  line 
is  simple.  For  the  112.5  mile  line  the  charging  current 
will  be  about  one  half  as  much  as  for  a  225  mile  line. 
Since  the  line  is  only  half  as  long  this  smaller  charging 
curent  will  flow  through  only  half  the  inductance  so  that 
the  net  result  of  half  the  line  charging  current  and  half 
the  inductance  will  be  about  one  fourth  the  voltage 


A  TYPICAL  220  KV  PROBLEM 


159 


boosting  effect  due  to  line  charging  current.  On  the 
other  hand  the  line  impedance  will  be  only  half  as  grear, 
but  the  net  result  will  be  more  condenser  capacity  for 
the  short  line.  A  large  part  of  the  condenser  capacity 
is  required  for  neutralizing  the  lagging  reactor  com- 
ponent of  the  load. 

Auxiliary  constant  A,  as  previously  explained,  ac- 
counts for  the  effect  of  the  line  charging  curent  flowing 
through  the  impedance  of  the  circuit;  that  is,  the  volt- 
age boosting  effect  of  the  charging  current.  Thus  for 
the  1 12.5  mile  line  (Fig.  73)  a^  which  accounts  for  the 
line  charging  current  flowing  through  the  inductance 
of  the  circuit  is  near  unity  and  a„  near  zero,  but  for  the 
450  mile  line  a^  drops  to  0.594  and  a,  increases  to 
0.07508.  As  the  length  of  line  increases,  constant  A 
moves  the  line  impedance  triangle  to  the  left  and  raises 
its  toe  somewhat.     The  increased  line  impedance  and 


slightly  increased  current  at  the  receiving  end  increases 
the  size  of  the  line  impedance  triangle. 

The  curves  of  Fig.  74  show  the  relation  between 
phase  modifier  capacity  and  sending  end  voltage  for  dif- 
ferent receiving  end  loads  of  85  percent  lagging  power- 
factor  and  a  constant  load  voltage  of  220  kv.  It  is  inter- 
esting to  note  the  effect  of  distance  for  fixed  size  con- 
ductors upon  the  maximum  amount  of  power  which  can 
be  transmitted  over  a  circuit,  as  evidenced  by  the  load 
curves  bending  upward  as  the  line  length  increase.  It  is 
also  interesting  to  note  the  decrease  in  phase  modifiers 
leading  capacity  and  increase  in  phase  modifier  lagging 
capacity  as  the  line  becomes  larger,  as  evidenced  of  the 
load  curves  shifting  to  the  right.  The  curves.  Fig.  75, 
show  the  voltage  at  each  side  of  the  raising  transformer, 
corresponding  to  various  condenser  capacities  in  paral- 
lel with  a  constant  load  of  75  000  kw  at  85  percent  lag- 
ging power- factor  and  220  kw. 


H.  B.  DWIGHT'S  METHOD. 


In  the  various  methods  for  determining  the  per- 
formance of  transmission  lines  which  are  described 
above,  current  and  voltage  vectors  or  corresponding 
vector  quantities  have  been  employed  throughout.  It 
was  believed  that  solutions  embodying  the  use  of  current 
and  voltage  vectors  would  be  the  more  easily  followed  by 
the  young  engineer,  for  the  assistance  of  whom  this  book 
has  been  primarily  written. 

H.  B.  Dwight  worked  out  and  published  in  book 
form  formulas  for  determining  the  complete  performance 
of  circuits  by  the  employment  of  quantities  not  gener- 
ally employed  in  the  methods  described  above.  These 
quantities  require  a  new  set  of  symibols  applicable  to 
his  method.  Partly  to  prevent  confusion  in  symbols  but 
principally  because  his  method  has  been  so  completely 
and  clearly  set  forth  and  illustrated  with  numerous  ex- 
amples worked  out  in  the  two  books  referred  to  his 
method  has  not  been  detailed  in  this  book.  To  include 
it  here  would  simply  be  a  duplication  of  what  is  al- 
ready available  in  very  complete   form. 

THE    CIRCLE    DIAGRAM 

Various  forms  of  circle  diagrams  as  an  aid  in 
determining  the  performance  of  short  transmission  lines 
have  been  frequently  described  by  writers,  notably  by 
R.  A.  Philip  'thru  the  medium  of  the  A.  I.  E.  E.  trans- 
actions of  February  1911.  Following  this  H.  B.  Dwight 
worked  out  a  solution  and  construction  for  a  circle 
diagram  which  accurately  takes  into  account  the  effect 
of  capacitance  in  transmission  lines  that  is,  a  circle 
diagram  for  long  high  voltage  lines.  This  circle  dia- 
gram consists  of  curves  which  indicate  the  phase  modi- 


fier capacity  (leading  or  lagging)  required  to  maintain 
a  certain  reciving  end  voltage  corresponding  to  all  values 
of  delivered  load  up  to  the  maximum  capacity  of  the 
line.  In  other  words  it  gives  data  such  as  is  given 
by  the  curves  of  Fig.  No.  70. 

The  next  step  in  the  development  of  the  circle  dia- 
gram was  to  so  alter  the  constants  upon  which  it  is  con- 
structed that  it  will  take  accurately  into  account  the 
localized  impedance  and  loss  in  raising  or  lowering 
transformers  or  in  both.  Of  course^  the  transformer 
impedance  may  be  added  to  the  line  impedance  as  is 
frequently  done  and  considered  as  distributed  line  im- 
pedance; Such  procedure,  will,  however,  in  the  case  of 
the  circle  diagram  for  the  line  alone  result  in  objection- 
able errors  in  the  results.  In  order  to  correctly  apply 
the  circle  diagram  to  long  lines  so  as  to  accurately  in- 
clude the  effect  of  transformers  in  the  circuit  it  is  neces- 
sary to  develop  new  formulas  for  obtaining  values  for 
the  constants  by  which  the  circle  diagram  is  constructed. 
See  articles  on  transmission  line  constants  by  R.  D. 
Evans  and  H.  K.  Sels  in  the  Electric  Journal,  page  306 
July  1921,  page  356  August  1921  and  page  530  Decem- 
ber 1921. 

To  the  exi>ert  who  spends  much  time  investigating 
transmission  problems  the  general  use  of  the  circle  dia- 
gram should  be  of  great  assistance.  It  indicates  per- 
formance at  all  loads,  which  with  other  methods  would 
hav^e  to  be  obtained  by  a  separate  calculation  or  vector 
diasfram  construction  for  each  load. 


♦Transmission  Line  Formulas,  1913.  D.  Van  Nostrand  Co- 
New  York  City  and  Constant-voltage  Transmission,  1915,  John 
Wiley  &   Sons   Inc.,  New   York  City. 


INDEX 


Pages. 

Accuracy  of  Various  Methods — Comparative  119  &  140 

Admittance   Correcting   Factor   for  Kennelly  Equivalent 

V   Solution— Charts  XX  and  XXI 100-101 

Advantages  of  High  Power  Factor 134 

Angle — Hyperbolic    88 

Application  of  Tables  to  the  Solution  of  Short  Lines — 

Chart  II  52 

Long    Lines— Chart    VIII 70 

Armature  Current — Effect  Upon  Field  Excitation  of  A. 

C.   Generators   129-130 

Armature    Impedance — Effect    Upon    Voltage   of   A.    C. 

Generators    130 

Auxiliary  Constants — Corresponding  to  Localized  Capa- 
citance methods   1 12-1 13 

Comparison  of  Results  When  Taken  from  Wilkinson 

and  Kennelly  Charts  146-lSl 

Definition  of 64-142 

Determination  by   Convergent  Series 80  &  82 

Determination  by   Hyperbolic  Functions 96 

Five   Different   Methods  of   Determining 143 

How  They  Modify  Short  Line  Diagrams 143 

Tabulation  for  64  Different  Circuits 83 

Wilkinson  Chart  for  Obtaining  A — Chart  V 67 

Wilkinson  Chart  for  Obtaining  B— Chart  VI.../. 68 

Wilkinson  Chart  for  Obtaining  C— Chart  VII 69 

Behavior  of  A.  C.  Generators  when  Charging  a  Trans- 
mission  Line 136 

Bibliography  on  Solution  of  Circuits 109 

On    Cable 128 

Cables — Capacitance  of  3-Conductor 125-126 

Capacitance  and  Susceptance  of  3-Conductor — Table 

XXVir   126 

Charging  Kv-a  for  3-Conductor— Table  XXVIII 127 

Effect  of  Stranding  and  Spiraling  Upon  Inductance  9 

Heating  Limits 121-122 

Inductance,    Reactance,    Impedance    of    3-Conductor 

Cable  at  25  Cycles— Table  XXV 123 

Inductance,    Reactance,    Impedance    of    3-Conductor 

Cable  at  60  Cycles— Table  XXVI : 124 

Reactance  of  3-Conductor   Cable 121 

Capacitance — Charging  Current,   Inductance — Chapter   II  10 

Definition   of H 

Formula  11  &  20 

Relation  to  Inductance 21 

Susceptance  Bare  Conductors  at  25  Cycles — Table  IX  17 

Susceptance  Bare  Conductors  at  60  Cycles — Table  X  18 

Three    Conductor    Cable 125-126 

To  Neutral  per  1000  Feet  of  Single  Bare  Conductor 

—Table    VIII 16 

Capacity    of    Synchronous    Motor    and    Condensers    for 

Power  Factor  Improvement 132 

of  Phase  Modifiers  for  Voltage  Control  Chapter  XV  138 
of    Bare    Conductors   in   air    (heating   limit)    Table 

XXIII  43 

Susceptance  to  Neutral  Per  Mile  of  Single  Bare  Con- 
ductor   at  25    Cycles— Table    IX 17 

Susceptance  to  Neutral  Per  Mile  of  Single  Bare  Con- 
ductor at  60  cycles — Table  X 18 

Charging  Current — At  Zero  Load 20 

Capacitance,   reactance — Chapter   II 10 

Effect  upon  Conductor  Loss 34 

Of    Short   Lines 62 

Relation  in  Single  and  Three  Phase  Circuits 20 

Charging   Kv-a — In   three    Phase   Circuits    Per    Mile   of 

3   Bare  Conductors — Table  XI 19 

of   3-Conductor   Cables 127 

Charging  Transmission  Lines — Behavior  of  A.   C.  Gen- 
erators   when 136 

Chart  I     Inductance   6 

11    Application  of   Tables   to   Short   Transmission 

Lines    52 

III  Mershon  Chart  for  Determining  Line  Drop  in 
Short  Lines  54 

IV  Dwight  Chart   for   Determining  Line  Drop   in 
Short    Lines 56 


Pages. 

V     Wilkinson  Chart  A  for  Determining  Auxiliary 

Constants  a.^  and  aj 67 

VI     Wilkinson  Chart  B   for  Determining  Auxiliary 

Constants  K  and  h, 68 

VII  Wilkinson  Chart  C  for  Determining  Auxiliary 
Constants  Ci  and  Cj 69 

VIII  Applicatiort  of    Tables   to   Long   Transmission 
Lines  70 

IX     Peters  Efficiency  Chart  for  Transformers 74 

X     Peters  Regulation  Chart  for  Transformers 75 

XI    Determination  of  Auxiliary  Constants  by  Con- 
vergent   Series 82 

XII    Auxiliary  Constants  of  64  Different  Circuits 83 

XIII  Calculation    of    Performance     (Receiver    End 
Conditions    Fixed) 84 

XIV  Calculation     of     Performance     (Sending     End 
Conditions    Fixed) 86 

XV     Calculated  Performance  of  64  Different  Circuits  87 
XVI     Determination  of  Auxiliary  Constants  by  Hy- 
perbolic   Functions 96 

XVII     Equivalent  'V'  Solution  of  Problem  X 103 

XVIII     Kennelly  Chart  for  Impedence  Correcting  Fac- 
tor   (Angles  0  to   .40) 98 

XIX     Kennelly  Chart  for  Impedance  Correcting  Fac- 
tor  (Angles  .40  to  1.0) 99 

XX     Kennelly  Chart  for  Admittance  Correcting  Fac- 
tor   (Angles  0  to  .20) 100 

XXI     Kennelly  Chart  for  Admittance  Correcting  Fac- 
tor  (Angles  .20  to  .50) 101 

XXII     Comparison  of  Results  by  Various  Methods 118 

XXIII    220    Kv.    Problem— Normal    Load     (Complete 

Solution)    149 

XXIV    220    Kv.    Problem— Normal    Load    (Approxi- 
mate   Solution) 150 

XXV    220    Kv.    Problem— Emergency    Load     (Com- 
plete   Solution) 153 

XXVI     220  Kv.  Problem — Emergency  Load  (Approxi- 
mate   Solution) 154 

XXVII  220  Kv.   Problem— Zero   Load    (Complete   So- 
lution)      155 

XXVIII  220    Kv.    Problem— Zero    Load    (Approximate 
Solution)  156 

Checking  the   Work 138 

Choice  of  Various  Methods  108 

Circuits — Paralleling 41 

Electric,  Dielectric  and   Magnetic 1-2 

Circular    Functions 88 

Sines,  Cosines,  Tangents — Table  K 57 

Common  Transmission  Voltages — Table  H 48 

Comparison — Accuracy   of   9    Different    Methods 118-119 

Accuracy  of  5  Methods  of  Including  Transformers....  140 
of  Calculated  Capacitance  of  3-Conductor  Cables  with 

test   results 128 

of  9  Different  Methods— Chapter  XII Ill 

Complex    Angle — Definition 90 

Complex  Hyperbolic  Functions — Definition 90-94 

Dr.  Kcnnelly's  Model 92 

Complex    Quantities — Definition 78 

Condensers  and  Phase  Modifiers 131 

Condensers  and  Synchronous  Motors — For  Power  Factor 

Improvement — Chapter    XIV 129 

Determination    of    Capacity 132-134 

Condenser.? — Generators    as 131 

Installations  of  Large  Capacity — Table  U 137 

Location    132 

Methods  of  applying — Chapter  XII Ill 

Mechanical    Load    Carried 131 

Ratings,   Starting,   "\"'  Curves,   Losses,   Etc 131 

Conductors — Capacitance   (See  Capacitance) 

Economic  Size 45  &  145 

Effect  of  Unsymmetrical  Spacing 10 

Effect  of   Stranding  and  Spiraling  upon   Inductance  9 

Flux   Distribution   Around 3 

Heating  of  Bare  Conductors  in  Air 42-43 

Heating  of  3-Conductor  Cables 121-122 

Inductance    (See    Inductance) 
Impedance   (See  Impedance) 


INDEX 


161 


Pages. 

Reactance  (See  Reactance) 
Resistance  (See  Resistance) 
Skin  Effect  (See  Skin  Effect) 

Weight  of  Bare— Table  E-1 46 

Constants — Auxiliary  (See  Auxiliary  Constants) 

Methods  of  Determining  the  Linear  Constants 51 

Transformer  Constants  Taken  Into  Account 7i  &  139 

Convergent  Series  for  Determining  the  Auxiliary  Constants  80-82 

Copper  Loss  of  Transformers — Table  X 141 

Copper  Conductors   (See  Conductors) 

Corona — Effect   of — Chapter    IV 35 

Formulas    , 36-37 

Voltage  Limitation— Table  XXII 38 

Correcting  Factors — Charts  for  Impedance  for  Equivalent 

T  Solution  Charts  XVIII  and  XIX 98-99 

Charts  for  Admittance  for  Equivalent  ■^  Solution^ 

Chart    XX    and    XXI 100-101 

Mathematical  Determination  for  Equivalent  tt  Solu- 
tion          107 

Cosines,  Sines  and  Tangents  of  Angles — Table  K 57 

Cost — Relative   Cost   of   High   Tension   Apparatus 46 

Power  Factor  Improvement  by  Synchronous  Motors  135 

Current  and   Voltage   Determination  Along  Circuit — By 

Auxiliary  Constants 62-64  &  76 

by  Hyperbolic   Position  Angles 102-106 

Degree — Subdivisions  of,  Table  P 110 

Determination  of — Capacity  of  Synchronous  Motors  and 

Condensers    132-134 

Correcting  Factor  for  Equivalent  tt  Solution  Mathe- 
matically            107 

Frequency  and  Voltage 45 

Dielectric  Circuit 2 

Distribution  of  Current  and  Voltage  Along  the  Circuit — 

By  Auxiliary  Constants 62-64  &  76 

by  Hyperbolic  Position  Angles 102-106 

by  a  Polar  Diagram  108 

Dwight  Chart  for  Short  Lines 55-56 

Economies  Size  of  Conductors 45  &  145 

Effect  of  Armature  Current  Upon  Field  Excitation  of 

A.  C.  Generators 129-130 

Armature    Impedance    upon    Terminal    Voltage    of 

A.  C.  Generators  130 

Corona — Chapter    IV 35 

Charging  Current  on  Conductor  Loss 34 

Harmonies  in  Current  and  Voltage 108 

Power  Factor  Upon  Voltage  Drop 135 

Spiraling  and  Stranding  of  Conductors  Upon  Induc- 
tance      9 

Transformers  in  the  Line 7i  &  139 

Corona — Chapter    IV 35 

Effective  Spacing  of  Conductors 10 

Efficiency  Chart  for  Transformer — Peters  Chart  IX 74 

Electric  Circuit — The  2 

Electric  Propogation — Speed  of 40 

Electric  Wave— Length  of 40 

Equivalent  tt  Method — Charts  for  Impedance  Correcting 

Factor— Charts  XVIII  &  XIX 98-99 

Charts    for    Admittance    Correcting    Factor — Charts 

XX  &  XXI 100-101 

General   97 

Example  of  Calculation— Chart  XVII 103 

Mathematical  Determination  of  the  Correcting  Fac- 
tors          107 

Equivalent  "T"  Solution— General  102 

Estimating   Tables — Quick — Chapter   III 23 

Exciting  Transmission  Lines — Methods  of 136 

Field  Excitation — Effect  of  Armature  Current  Upon 129-130 

Flux — Effect  of  Armature  Flux  Upon  Field  Excitation 

of  A.  C.  Generator 129-130 

Distribution  Around   Conductor 3-6 

Formulas 7  &  9 

Formulas — Capacitance  11   &  20 

Carrying  Capacity  of   Bare  Conductors  in  Air 44 

Convergent   Series 80 

Corona    36-37 

Inductance  7  &  9 

Transmission  Line  in  Terms  of  the  Auxiliary  Con- 
stants     80 

Transmission  Lines  of  Short  Length 59 

Transmission  Line  in  Terms  of  Hyperbolic  Functions  80 

Frequency    Determination 45 


Pages. 

FuiKtions— Complex  Functions  of  Hyperbolic  Angles 90-94 

Circular  88 

Hyperbolic— Real  88 

Hyperbolic — Complex  91 

Sines,  Cosines  and  Tangents  of  Circular  Angles 57 

Generators — As   Synchronous  Condensers 131 

Behavior  When  Charging  Transmission  Lines 136 

Effect  of  Field  Excitation  Upon  A.  C.  Generators....l29-130 

Graphical   Solution    (See  Solutions) 

Harmonic  Currents  and  Voltages — Effect  of 108 

in  Quarter   Wave   Resonance 41 

Heating — Bare   Conductors   in   Air 42-43 

Limits   for  Cables,  General 121-122 

Tabulated  Values  for  Cables— Table  XXIV 122 

High   Power   Factor — Advantages 134 

High  Tension  Apparatus — Relative  Cost — Table  F 46 

How  High  to  Raise  the  Power  Factor 132 

Hyperbolic    Angles — Real 88 

Complex 9(^-94 

Hyperbolic  Functions — Applied  to  the  Solution  of  Line 

Performance — Chapter    XI 95 

Chapter    X 88 

Complex    Angles 91 

Formulas  for  Long  Lines 80 

for  Determining  the  Auxiliary   Constants 96 

Symbols  Pertaining  to 95 

Inductance — Capacitance — Charging  Current — Chapter  II  10 
Effect   of    Spiraling   and    Stranding   of    Conductors 

Upon  9 

Formulas  7  &  9 

(General 3 

Graphic    Solution — Chart    1 6 

Per  1000  Feet  of  Single  Conductor— Table  III 8 

Reactance  and  Impedance  of  3-Conductor  Cables  at 

25    Cycles— Table    XXV 123 

Reactance  and  Impedance  of  3-Conductor  Cables  at 

60    Cycles— Table    XXVI 124 

Relation  to  Capacitance 21 

Skin  Effect  and   Resistance — Chapter   1 1 

Variation  from  the  Fundamental  Formula 7 

Installations  of  Large  Phase  Modifiers — Table  U 137 

Impedance — Correcting    Factor    for   Equivalent  tt   Solu- 
tion—Charts XVIII  and  XIX 98  &  99 

Effect  of  Armature  Impedance  Upon  Voltage  of  A. 

C.    Generators „....  130 

Effect  of  Transformer  Impedance  in  the  Circuit 73  &  139 

Inductance    and    Reactance    of    3-Conductor    Cables 

at  25  Cycles— Table  XXV 123 

Inductance    and    Reactance    of    3-Conductor    Cables 

at  60  Cycles— Table  XXVI 124 

Transformer  Impedance  to  Neutral 141 

Iron  Loss  in  Transformers — Table  X 141 

Kennelly— Charts  for  Impedance  Correcting  Factors  for 

Equivalent  ,r  Solution— Charts  XVIII  and  XIX  98-99 

Charts  for  Admittance  Correcting  Factors  for  Equiva- 
lent T  Solution— Charts  XX  and  XXI lOO-lOl 

Equivalent  ,r  Solution — General 97 

Equivalent  ,r  Solution— Example  of  Solution— Chart 

XVII   103 

Model  for  Explaining  Functions  of  Complex  Angles  92 

Light  Speed — Relation  to  Inductance  and  Capacitance 21 

Location  for  Synchronous  Condensers 132 

Localized  Capacitance  Methods — Chapter  XII Ill 

Auxiliary   Constants   Corresponding   to 113 

Losses  in  Transformers — Table  X „ 141 

Synchronous    Condensers 131 

Magnetic  Circuit — The 2 

Magnetizing  Current  of  Transformers 141 

Mechanical  Load  Carried  by  Synchronous  Condensers....  131 

Mershon  Chart — General 53 

Mershon  Chart — Chart  III 54 

Where   it   Falls   in  Error    When   Applied   to   Long 

Lines  72  &  143 

Methods  of  Exciting  Transmission  Lines 136 

Solution   (See  Solution) 

Middle  Condenser  or  Nominal  T  Solution 115 

Model    for   Explaining  Complex    Functions  of  Complex 

Hyperbolic    Angles 92-94 


162 


INDEX 


Pages. 

Nominal  'V"  or  Split  Condenser  Solution 97  &  113 

"T"  or  Middle  Condenser  Solution 11 S 

Paralleling  Transmission  Lines 41 

Peeks   Corona  Formulas 36 

Performance  of  Short  Lines — Chapter  VII 49 

Composite    Lines 50 

Formulas    for 59 

Graphical  Methods 52 

Mathematical  Methods 57 

Procedure  in  Determining — Chart  II 52 

Performance  of  Long  Lines — By  Hyperbolic  Functions — 

Chapter   XI 95 

by   Convergent   Series — Chapter   IX 77 

by  Graphical  Method — Chapter  VIII 61 

by  Localized  Capacitance  Methods— Chapter  XII Ill 

Equivalent  jr  Method 97 

Procedure  in  Determining — Chart  VIII 70 

Tabulated  Performance  of  64  Circuits — Chart  XV....  87 

Typical  220  Kv  Problem— Chapter  XVI 145 

Performance  of  Long  Lines  Including  Transformers 145 

Peters   Efficiency  Chart   for  Transformers — Chart  IX....  74 

Regulation  Chart  for  Transformers — Chart  X 75 

Phase  Modifiers  for  Voltage  Control — Chapter  XV 138 

and    Synchronous    Condensers 131 

Installations  of 137 

Curves  of 152 

Capacity  of 155 

Polar  Diagram  of  Voltage  and  Current  Distribution  for 

Problem  X 108 

Position  Angles — Explanation  of 102-107 

Mathematical    Determination 107 

Solution  for  Voltage  and  Current  Along  the  Circuit 

by  Hyperbolic 102-106 

Power  Factor — Advantages  in  High   Power  Factor 134 

Cost  of  Improvement 135 

Examples  of  Determination  of  Improvement 133 

Effect  on  Voltage  Drop 135 

How  High  to  Raise 132 

Improvement  by  Synchronous  Motors  and  Condensers 

—Chapter   XIV 129 

Propogation — Speed  of  Electric 40 

Quick    Estimating 23-33 

Quarter   Wave    Resonance 40 

Quantities — Complex 78 

Ratings  of  Synchronous  Condensers 131 

Ratio  of  Reactance  to  Resistance  at  25°C.  and  25  Cycles 

—Table    VI 14 

60  Cycles— Table   VII 15 

Reactance  and  Resistance  of  Copper  and  Aluminum  Con- 
ductors at  25  Cycles,  Table  IV 12 

60  Cycles— Table  V 13 

A  High  Reactance  Problem 60 

Reactance — Capacitance    and    Charging    Current — Chap- 
ter II 10 

Inductance  and  Impedance  of  3-Conductor  Cables  at 

25    Cycles— Table    XXV 123 

Itiductance  and  Impedance  of  3-Conductor  Cables  at 

60  Cycles,  Table  XXVI 124 

Ratio  of  Reactance  to  Resistance  at  25°   C.  and  25 

Cycles— Table    VI 14 

Ratio  of  Reactance  to  Resistance  at  25°   C.  and  60 

Cycles— Table    VII 15 

Three-Conductor    Cables 121 

Transformers 141 

Regulation  Chart — Dwights — Chart  IV 56 

Mershons — Chart   HI 54 

Peters  Transformer— Chart  X 75 

Relation  of  Inductance  to  Capacitance 21 

Inductance  and  Capacitance  to  Speed  of  Light 21 

Charging  Current  in  Single  and  Three  Phase  System  20 

Relative  Cost  of  High  Tension  Apparatus — Table  F 46 

Resistance — Copper    Conductors — Genera! 2 

Copper  Conductors — Per  1,000  Feet — Table  1 4 

Copper  Conductors — Per  Mile — Table  II 5 

and  Reactance  of  Copper  and  Aluminum  Conductors 

—25   Cycles— Table   IV 12 

and  Reactance  of  Copper  and  Aluminum  Conductors 

60  Cycles— Table  V 13 

Ratio  Reactance  at  25  Cycles  to— Table  VI 14 

Ratio  Reactance  at  60  Cycles  to— Table  VII 15 

Transformers    141 

Resonance — Guarter    Wave 40 


Paces. 

Self  Induction — Effect  Upon  Voltage  Regulation 62 

Short   Lines — Formulas 59 

Performance    49 

Symbols    50 

Single  End  Condenser  Method 112 

Sines,  Cosines  and  Tangents  of  Circular  Angles 57 

Skin  Effect — In  Conductors 2 

Resistance  and  Inductance — Chapter  1 1 

Tabulation  of  Increase  in  Resistance  Due  to — Table  B  3 

Solution— A  Typical  220  Kv.  Problem 145 

Choice   of 108 

Comparison  of    Various   Methods — Chapter  XII Ill 

Comparative  Accuracy  of  Various  Methods 118-119 

Comparison  of  Short  and  Long  Line  Diagrams 72  &  142 

Complete  for  Long  Lines  Including  Transformers....  145 

Dwight   Chart 55-56 

Equivalent    ,r — General    97 

Equivalent  t, — Exainple — Chart  XVII 103 

Equivalent  T 102 

Graphical   vs.   Mathematical 51 

Graphical  for  Short  Lines 52 

Graphical  for  Problem  "X" 70 

Graphical  for  Long  Lines  Including  Transformers....  145 

Hyperbolic  Functions — Chapter  XI 95 

Localized   Capacity — Chapter   XII Ill 

Mershon  Chart 53-54 

Middle  Condenser  or  Nominal  T  Method 115 

Nominal  tt  or  Split  Condenser  Method 97  &  113 

Position    Angles 102-106 

Single  End  Condenser  Method 112 

Split  Condenser  or  Nominal  ^  Method 113 

Three  Condenser  or  Dr.  Steinmetz's  Alethod 116 

Typical  220  Kv  Problem— Chapter  XVI •      145 

Spacing   of   Conductors — Equivalent 10 

Split  Condenser  or  Nominal  ,r  Solution 113 

Speed  of  Electric  Propogation 40 

Speed  of  Light — Relation  of  Inductance  and  Capacitance 

to    21 

Steinmetz's  Three   Condenser   Method 116 

Sub-division  of  a  Degree — Table  P 110 

Susceptance — Three   Conductor   Cables — Table   XXVII....  126 

Overhead  Conductors  at  25  Cycles^Table  IX 17 

Overhead  Conductors  at  60  Cycles — Table  X 18 

Symbols — Corona  36 

Line   50 

Hyperbolic   95 

Synchronous  Condensers — (See  Condensers) 

Table  I    Resistance   of    Copper    Conductors   at   Various 

Temperatures  per   1000   feet 4 

II     Resistance   of    Copper    Conductors   at    Various 

Temperatures  Per  Mile 5 

III  Inductance  of  Single  Conductors  per  1000  feet  8 

IV  Resistance  and  25  Cycle  Reactance  per  Mile....  12 
V     Resistance  and  60  Cycle  Reactance  Per  Mile 13 

VI     Ratio  of  25  Cycle  Reactance  to  Resistance  at 

25°    C 14 

VII     Ratio  of  60  Cycle  Reactance  to  Resistance  at 

25°    C. 15 

VIII     Capacitance  of  Single  Conductor  per  1000  feet  16 
IX    25  Cycle  Capacity  Susceptance  of  Single  Bare 

Conductors    Per    Mile 17 

X    60  Cycle  Capacity  Susceptance  of  Single  Bare 

Conductors    Per    Mile 18 

XI     Charging  Kv-a,  3  Phase  Circuits,  Bare  Conduc- 
tors   Per   Mile 19 

XII     Quick  Estimating  Table  for  220  and  440  Volts  24 

XIII     Quick  Estimating  Table  for  550  and  1100  Volts  25 
XIV    Quick    Estimating    Table    for    2200,   4000    and 

4400    Volts 26 

XV    Quick  Estimating  Table  for  6000,  6600,   10000 

and    11000   Volts 27 

XVI     Quick  Estimating  Table  for  12,000,  13,200,  15,000 

and  16,500  Volts 28 

XVII    Quick  Estimating  Table  for  20,000,  22,000,  30,000 

and  33.000  Volts 29 

XVIII     Quick  Estimating  Table  for  40,000,  44,000,  50.000, 

and  60,000  Volts 30 

XIX     Quick    Estimating    Table    for    66,000,    70,000, 

80,000  and  88,000  Volts 31 

XX     Quick   Estimating   Table    for    100,000,    110,000, 

120,000,  132,000  and  140,000  Volts 32 

XXI     Quick    Estimating    Table    for    154,000,    187,000 

and  220,000  Volts 33 

XXII    Approximate  Voltage  Limitations  from  Corona  38 


INDEX 


163 


Pages. 

XXIir    Heating  Capacity  for  40°   C.  Rise  Bare  Con- 
ductors      43 

XXIV    Carrying  Capacity  of   Insulated  Copper  Con- 
ductors    122 

XXV    Inductance,      Reactance,      Impedance      at      25 

Cycles,  3-Conductor  Cables 123 

XXVI    Inductance,  Reactance,  Impedance  at  60  Cycles 

3-Conductor   Cables 124 

XXVII  Capacitance   and   Susceptance   Per   Mile  of  3- 
Conductor   Paper  Insulated  Cables 126 

XXVIII  Charging    Kv-a    of    3-Conductor    Cables    Per 

Mile    127 

Tangents,  Sines  and  Cosines  of  Circular  Angles  (Table 

K)   57 

Three  Condenser  or  Dr.  Steinmetz  Method 116 

Transformers — Constants    141 

Effect  in  Circuit 73  &  139 

Iron  Loss,  Impedance,  Magnetizing  Current 141 

Peters    Efficiency    Chart 74 

Peters    Regulation    Chart 75 

Reactance,  Resistance 141 

Transmission  Lines — Behavior  of  A.  C.  Generators  When 

Charging    136 

Common   Voltages 48 


Paces. 

Excitation  of 136 

Paralleling 41 

Performance   (See  Performance  and  Solutions) 

Determination  of  Frequency  and  Voltage 45 

Typical  220  Kv  Problem 145 

Symmetrical  Spacing  of  Conductors 10 

Various  Methods — Comparison  of Ill 

Variation  of  Current  and  Voltage  Along  the  Circuit  62-64  &  76 

Vector   Operations — General 78-80 

As  Applied  to  Problem  "X" 85 

"V"  Curve  of  Synchronous  Condenser 131 

Voltage — Control  by  Means  of  Phase  Modifiers — Chap- 
ter XV  138 

Common    Transmission 48 

Distribution  Along  the  Circuit 62-64  &  76 

Effect  of  Armature  Impedance  Upon  Generator  Volt- 
age     130 

Limitations  Due  to  Corona  Effect — Table  XXII 38 

Determination  of 45 

Wave    Length 40 

Wilkinson  Charts— Charts  A,  B  and  C  for  the  Auxiliary 

Constants    67-69 

Weight  of  Bare  Copper  Conductors 46 


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